Okay, so I need to figure out the shortest path a gecko can take to reach the fly in this room. Let me start by visualizing the room. The room is 12 feet long, 10 feet wide, and 8 feet tall. The gecko is on one of the side walls, which is 10 feet by 8 feet. It's one foot from the ceiling and one foot from the back wall. The fly is on the opposite side wall, which is the other 10 feet by 8 feet side. The fly is one foot from the floor and one foot from the front wall. 

The gecko can only walk across the ceiling and walls, not jump. So, it's confined to moving along the surfaces. To find the shortest path, I remember that for such problems, unfolding the room into a 3D net can help visualize the shortest path as a straight line in the unfolded 2D plane. 

First, let's label the walls and positions. The gecko is on the back wall (since it's one foot from the back wall and one foot from the ceiling). The fly is on the opposite side wall, which would be the front wall? Wait, no. Wait, the room has two side walls (each 10x8), a front wall (12x8), and a back wall (12x8). But the gecko is on a side wall, which is 10 feet wide and 8 feet tall. The fly is on the opposite side wall. So, opposite side walls are the two 10x8 walls. So, the gecko is on the back side wall (12x8), but wait, the side walls are 10x8. Wait, maybe I need to clarify the room dimensions.

Wait, the room is 12 feet long, 10 feet wide, and 8 feet tall. So, the walls are as follows: the front and back walls are 12 feet long and 8 feet tall (since length is 12, width is 10, so front and back walls are 12x8). The side walls are 10 feet long and 8 feet tall (width is 10, so side walls are 10x8). 

So, the gecko is on a side wall (10x8). It's one foot from the ceiling (so 7 feet from the floor) and one foot from the back wall. Wait, the back wall of the side wall? Wait, maybe the gecko is on the side wall, which is 10 feet wide (along the width of the room) and 8 feet tall. So, the back wall of the side wall would be the 12-foot-long wall? Hmm, maybe I need to better visualize this.

Let me sketch the room in my mind. The room is length 12 (front to back), width 10 (left to right), and height 8 (floor to ceiling). The walls are: front (12x8), back (12x8), left (10x8), right (10x8), ceiling (12x10), and floor (12x10). 

The gecko is on the left side wall (10x8), one foot from the ceiling (so 7 feet above the floor) and one foot from the back wall (12x8). Wait, the back wall is 12 feet long. If the gecko is on the left side wall (which is 10 feet wide), being one foot from the back wall would mean it's one foot away from the wall that's 12 feet long. So, the left side wall is adjacent to the back wall? Wait, no. The left side wall is 10 feet wide (along the width of the room), and the back wall is 12 feet long (along the length of the room). So, the left side wall is adjacent to the front and back walls. Wait, maybe the coordinates can help.

Let me assign coordinates to the room. Let's consider the room with coordinates (x, y, z), where x is from 0 to 12 (length), y is from 0 to 10 (width), and z is from 0 to 8 (height). The four walls on the sides are at x=0 (left), x=12 (right), y=0 (front), y=10 (back). The ceiling is z=8, floor is z=0.

The gecko is on the left wall (x=0), one foot from the ceiling (z=7) and one foot from the back wall (y=9). Wait, the back wall is at y=10. So, the gecko is at (0, 9, 7). The fly is on the opposite side wall. The opposite side wall to the left wall (x=0) is the right wall (x=12). The fly is one foot from the floor (z=1) and one foot from the front wall (y=1). So, the fly is at (12, 1, 1).

Now, the gecko needs to move from (0,9,7) to (12,1,1) along the surfaces. The surfaces it can traverse are the walls and ceiling. So, it can't go through the air or the floor. So, the path must be along the walls, ceiling, and possibly the ceiling and other walls.

To find the shortest path on the surfaces, we can use the method of reflecting the room into a plane. By unfolding the walls and ceiling into a flat net, the shortest path becomes a straight line in this net. The key is to find the correct unfolding that connects the gecko's starting position to the fly's position.

Let me think about how to unfold the walls. The gecko is on the left wall (x=0). The fly is on the right wall (x=12). To connect these two points, we can unfold the room by reflecting across the ceiling or the adjacent walls.

Since the gecko is at (0,9,7) and the fly is at (12,1,1), the direct straight line through the interior would pass through the floor, which is not allowed. So, we need to find a path that goes along the surfaces.

One approach is to consider the path going from the gecko's position, moving up to the ceiling, then moving along the ceiling to the fly's position. Alternatively, it could go down along a wall, then up another wall. But we need to find the shortest such path.

Alternatively, we can model the 3D room as a series of connected 2D walls and unfold them into a plane. For example, if we unfold the left wall (x=0) with the ceiling (z=8) and the right wall (x=12), the path can be represented as a straight line in this unfolded plane.

Let me try to visualize this. If we unfold the left wall (x=0) with the ceiling (z=8), then the gecko's position is at (0,9,7). If we unfold the right wall (x=12) into the same plane, the fly's position is at (12,1,1). But since we are considering moving along the surfaces, we need to reflect the fly's position across the ceiling or another wall to create a straight path.

Alternatively, maybe reflecting the coordinate system. Let me think: when we unfold the walls, we can represent the fly's position as a reflection of its actual position across the adjacent walls. For example, if the gecko is moving from the left wall to the right wall, we can reflect the fly's position across the ceiling or the adjacent walls to compute the straight-line distance.

Wait, perhaps a better way is to compute the distance in the unfolded net. Let me consider the gecko's position as (0,9,7). To get to the fly's position (12,1,1), the gecko can move along the left wall up to the ceiling, then move horizontally along the ceiling to the fly's position, then move down the right wall. Alternatively, it could move down along the left wall, then horizontally along the floor, then up the right wall. But the problem is that the gecko cannot move through the air, so it has to stay on the walls or ceiling.

Alternatively, we can create a net that includes the left wall, the ceiling, and the right wall. When unfolded, the left wall (x=0) and the right wall (x=12) are adjacent to each other. The gecko is on the left wall, one foot from the ceiling, so at (0,9,7). The fly is on the right wall, one foot from the front wall (y=1) and one foot from the floor (z=1). So, the fly is at (12,1,1).

If we unfold the left wall and the ceiling into a plane, the gecko's position is (0,9,7). The ceiling is 12x10, so when we unfold the left wall (which is 10x8) and the ceiling (12x10), the total height becomes 10 (width of the room) + 8 (height of the left wall). Wait, maybe not. Let me think again.

Alternatively, to create a net that includes the left wall, the ceiling, and the right wall. The left wall is 10x8, the ceiling is 12x10, and the right wall is 10x8. When we unfold the left wall and the right wall next to the ceiling, the total width becomes 10 (left wall) + 12 (ceiling) + 10 (right wall) = 32 feet? Wait, that seems too much. Maybe I need to adjust.

Wait, perhaps it's better to model the path as moving from the gecko's position to the fly's position by unfolding the walls into a plane. Let me consider the gecko moving up the left wall to the ceiling, then moving horizontally along the ceiling to the fly's position, then moving down the right wall. The total distance would be the sum of the vertical movement up the left wall, the horizontal movement along the ceiling, and the vertical movement down the right wall. However, since the gecko can't move through the air, the path must be along the walls and ceiling, so the straight line in the unfolded net would represent the shortest path.

Alternatively, think of the room as a 3D grid and compute the distance by unfolding the walls. Let me try to compute the coordinates in the unfolded net.

First, the gecko is at (0,9,7). The fly is at (12,1,1). If we unfold the left wall (x=0) and the ceiling (z=8) into a plane, we can represent the gecko's position as (0,9,7) and the fly's position as (12,1,1). Wait, but the fly is on the right wall, so when we unfold the right wall into the same plane as the left wall and ceiling, the fly's position would be reflected.

Alternatively, imagine unfolding the left wall (x=0) and the right wall (x=12) into a single plane. The gecko is at (0,9,7). The fly is at (12,1,1). But since we can't go through the air, we need to find a path that goes along the walls. So, maybe the shortest path goes up the left wall to the ceiling, then along the ceiling to the right wall, then down the right wall to the fly. The total distance would be the sum of the vertical movement up, the horizontal movement along the ceiling, and the vertical movement down.

Alternatively, the gecko can move down the left wall, then along the front wall, then up the right wall. But the fly is on the opposite side wall, so maybe that's not optimal.

Wait, maybe the key is to compute the shortest path by reflecting the room. Let's consider reflecting the fly's position across the ceiling and walls to create an image that can be connected by a straight line through the surfaces.

If we reflect the fly's position across the ceiling, it would be at (12,1,15) (since the ceiling is at z=8, reflecting across it would add 8*2 -1 =15? Wait, reflection across the ceiling (z=8) would invert the z-coordinate. So, the fly is at z=1, reflecting across the ceiling would make it z=8 + (8 -1) =15. So, the reflected position would be (12,1,15). Then, the straight line from the gecko's position (0,9,7) to the reflected fly's position (12,1,15) would cross the ceiling, which is at z=8. The intersection point would be where the gecko's path meets the ceiling.

Alternatively, reflecting the fly's position across the right wall (x=12). The reflection would be at (-12,1,1). Then, the straight line from (0,9,7) to (-12,1,1) would cross the right wall at x=12. But since the gecko is on the left wall, moving to the right wall, this reflection might not help.

Alternatively, maybe reflecting the fly's position across the ceiling and the right wall. Let me think. The gecko is at (0,9,7), and the fly is at (12,1,1). To create a straight path, we can reflect the fly's position across the ceiling (z=8) to get (12,1,15). Then, the line from (0,9,7) to (12,1,15) would cross the ceiling at some point. The distance would be the straight line distance in 3D, but since we're constrained to the surfaces, this line would correspond to the path on the surfaces.

Wait, but the problem is that the gecko can't pass through the ceiling; it has to walk on the walls and ceiling. So, the reflection method works when you can reflect across walls and ceilings to compute the shortest path.

Let me formalize this. The idea is that when you reflect the fly's position across the ceiling, the shortest path from the gecko's position to the fly's position via the ceiling is the straight line distance from the gecko's position to the reflected fly's position. But in our case, the gecko is on the left wall, and the fly is on the right wall. So, to connect them, we need to reflect the fly's position across the ceiling and the adjacent walls.

Wait, maybe the correct reflection is to reflect the fly's position across the ceiling and the right wall. Let me try that. The fly is at (12,1,1). Reflecting across the ceiling (z=8) gives (12,1,15). Then, reflecting across the right wall (x=12) would invert the x-coordinate, so (-12,1,15). Then, the straight line from (0,9,7) to (-12,1,15) would cross the right wall and the ceiling. The length of this line would be the distance in 3D, which corresponds to the path on the surfaces.

But maybe I need to adjust the reflection. Alternatively, perhaps the correct reflection is across the ceiling and the adjacent walls.

Wait, another approach: when the gecko moves from the left wall to the right wall, the path can be represented as moving along the left wall up to the ceiling, then along the ceiling to the right wall, then down the right wall to the fly. The total distance would be the sum of the vertical movement up, the horizontal movement along the ceiling, and the vertical movement down. But maybe this isn't the shortest path.

Alternatively, if the gecko could move diagonally across the ceiling and walls, the path might be shorter. For example, moving from the gecko's position on the left wall up to the ceiling, then diagonally across the ceiling to the fly's position on the right wall. But since the fly is below the ceiling, the gecko would have to move down the right wall after crossing the ceiling.

Wait, maybe the key is to consider the gecko's path going up the left wall, crossing the ceiling, then moving diagonally across the ceiling to the fly's position. But the fly is below the ceiling, so maybe that's not possible. Alternatively, the gecko could move down the left wall, cross the front wall, then move up the right wall. But the fly is one foot from the front wall (y=1) and one foot from the floor (z=1). Hmm, this is getting complicated.

Let me try to use the reflection method more carefully. The reflection method works by reflecting the destination point across the walls and ceilings that the source can traverse. Since the gecko can walk on the ceiling and the walls, we need to reflect the fly's position across the adjacent walls and ceiling to create a straight path.

In this case, the gecko is on the left wall (x=0), and the fly is on the right wall (x=12). The gecko can walk on the ceiling (z=8) and the walls (y=0 to 10). To find the shortest path, we can reflect the fly's position across the ceiling and the right wall. Let's see.

First, reflect the fly's position across the ceiling (z=8). The fly is at (12,1,1). Reflecting across the ceiling gives (12,1,15). Then, reflect this point across the right wall (x=12) to get (-12,1,15). Now, the straight line from the gecko's position (0,9,7) to (-12,1,15) would cross the right wall and the ceiling. The length of this line would be the distance in 3D, which corresponds to the path on the surfaces.

Calculating the distance between (0,9,7) and (-12,1,15):

Δx = -12 - 0 = -12

Δy = 1 - 9 = -8

Δz = 15 - 7 = 8

Distance = √[(-12)^2 + (-8)^2 + (8)^2] = √[144 + 64 + 64] = √272 = √(16*17) = 4√17

But wait, this distance is the straight line in the reflected 3D space. However, since the gecko is constrained to the surfaces, the actual path would be the same as this straight line, but on the unfolded surfaces. Therefore, this distance is the length of the shortest path the gecko can take.

But let me check if this makes sense. The gecko starts at (0,9,7). If we reflect the fly's position across the ceiling and the right wall, the straight line in the reflected space corresponds to the gecko's path going up the left wall, crossing the ceiling, moving diagonally across the ceiling to the reflected position, then down the right wall. However, since the gecko can't move through the air, the actual path is along the surfaces, but the distance is the same as the straight line in the reflected space.

Therefore, the shortest path length is 4√17.

Wait, but let me verify this with another approach. Suppose we unfold the left wall, the ceiling, and the right wall into a flat plane. The left wall is 10x8, the ceiling is 12x10, and the right wall is 10x8. When we unfold the left wall and the ceiling, the total height becomes 8 + 10 = 18 feet (from the left wall and the ceiling). Then, the right wall is 10 feet wide, so the total width would be 12 (ceiling) + 10 (right wall) = 22 feet? Wait, maybe not. Let me think again.

Alternatively, when we unfold the left wall and the ceiling, the gecko's position is at (0,9,7). The ceiling is 12x10, so if we unfold the left wall (10x8) to the left of the ceiling, the total width becomes 10 (left wall) + 12 (ceiling) = 22 feet. The gecko's position is at (0,9,7) on the left wall. The fly's position is on the right wall (12x8), which is 12 feet from the left wall. So, in the unfolded plane, the fly's position would be at (12 + 10, 1, 1) = (22,1,1)? Wait, no. Because when we unfold the ceiling, the fly's position is on the right wall, which is adjacent to the ceiling. So, the fly's position is at (12,1,1). When we unfold the ceiling to the left wall, the fly's position would be at (12,1,1) in the ceiling's plane. But since the ceiling is 12x10, we can represent the fly's position as (12,1,1). However, the left wall is 10x8, so the total unfolded width is 10 (left wall) + 12 (ceiling) = 22. The gecko is at (0,9,7) on the left wall, and the fly is at (12,1,1) on the ceiling. So, the straight line distance in this unfolded plane would be from (0,9,7) to (12,1,1). Wait, but the gecko is on the left wall (x=0), the fly is on the ceiling (z=8), but in the unfolded plane, the fly's position is (12,1,1). Wait, this is getting confusing.

Alternatively, perhaps the correct way is to consider the gecko's position (0,9,7) and the fly's position (12,1,1). To find the shortest path on the surfaces, we can reflect the fly's position across the ceiling and the right wall. As I did before, reflecting across the ceiling gives (12,1,15), then reflecting across the right wall gives (-12,1,15). The straight line distance from (0,9,7) to (-12,1,15) is √[(-12)^2 + (-8)^2 + (8)^2] = √[144 + 64 + 64] = √272 = 4√17. Therefore, the shortest path length is 4√17.

But let me check if there's another reflection that could result in a shorter path. For example, reflecting the gecko's position across the left wall or the ceiling. Wait, reflecting the gecko's position across the ceiling would give (0,9,15). Then, the fly's position is (12,1,1). The straight line distance would be √[(12)^2 + (-8)^2 + (14)^2] = √[144 + 64 + 196] = √404, which is larger than 4√17. So, that's longer. Therefore, the previous reflection gives the shorter path.

Alternatively, reflecting the fly's position across the right wall and the ceiling. Reflecting the fly's position (12,1,1) across the right wall (x=12) gives (-12,1,1). Then reflecting across the ceiling gives (-12,1,15). The straight line distance from (0,9,7) to (-12,1,15) is the same as before, √[(-12)^2 + (-8)^2 + (8)^2] = √272 = 4√17. So, same result.

Therefore, regardless of the order of reflections, the distance comes out to 4√17. Therefore, the shortest path length is 4√17 feet.

But let me double-check this. Let's compute the coordinates again. The gecko is at (0,9,7). The fly is at (12,1,1). If we reflect the fly's position across the ceiling (z=8), it becomes (12,1,15). Then, reflecting this across the right wall (x=12) gives (-12,1,15). The distance from (0,9,7) to (-12,1,15) is √[(-12)^2 + (1-9)^2 + (15-7)^2] = √[144 + 64 + 64] = √272 = 4√17. Yes, that's correct.

Alternatively, if we don't reflect across both walls and the ceiling, but only across one wall and the ceiling, the distance might be longer. For example, reflecting only across the ceiling, the fly's position becomes (12,1,15). The straight line from (0,9,7) to (12,1,15) is √[(12)^2 + (-8)^2 + (8)^2] = √[144 + 64 + 64] = √272 = 4√17. Wait, that's the same distance. So, maybe reflecting only across the ceiling is sufficient? Wait, but the fly is on the right wall, so to reach it, the gecko has to cross the right wall. So, reflecting across the ceiling and the right wall is necessary.

Alternatively, perhaps reflecting across the ceiling alone isn't enough because the gecko needs to reach the right wall, which is adjacent to the ceiling. Therefore, the reflection across both the ceiling and the right wall is needed. Therefore, the distance remains 4√17.

Therefore, the shortest path length is 4√17 feet. To confirm, let's compute this numerically. 4√17 ≈ 4*4.123 = 16.492 feet. Let's see if that makes sense. The gecko is 1 foot from the ceiling and back wall, so it's 7 feet down the wall. The fly is 1 foot from the floor and front wall, so 1 foot up from the floor. The distance through the ceiling would be the hypotenuse of a right triangle where one leg is the distance along the ceiling from the gecko's projection to the fly's projection, and the other leg is the vertical difference. Wait, but the gecko is on the left wall, so its projection on the ceiling would be (0,9,8). The fly's projection on the ceiling would be (12,1,8). The distance between (0,9,8) and (12,1,8) is √[(12)^2 + (-8)^2] = √(144 + 64) = √208 = 4√13 ≈ 14.422 feet. Then, the vertical difference between the gecko and the fly is 7 - 1 = 6 feet. So, the total distance would be √[(4√13)^2 + 6^2] = √(208 + 36) = √244 ≈ 15.62 feet. Wait, but that's less than 16.492. Hmm, that's conflicting with the previous result.

Wait, maybe I made a mistake here. Let's clarify. If we consider the gecko's position (0,9,7) and the fly's position (12,1,1). The straight line distance through the ceiling would be the hypotenuse of the horizontal distance between their projections on the ceiling and the vertical distance between them. The projections on the ceiling are (0,9,8) and (12,1,8). The horizontal distance is √[(12-0)^2 + (1-9)^2] = √(144 + 64) = √208 = 4√13. The vertical distance is |7 - 1| = 6. Then, the straight line distance through the ceiling would be √[(4√13)^2 + 6^2] = √(208 + 36) = √244 ≈ 15.62 feet. However, this is a straight line through the ceiling, which is not possible because the gecko can't pass through the ceiling. Therefore, this distance is not achievable. Therefore, the reflection method gives the correct distance of 4√17 ≈ 16.492 feet, which is longer than 15.62 feet, but since the path must stay on the surfaces, the 16.492 feet is indeed the shortest possible.

Therefore, the answer is 4√17 feet.

**Final Answer**
The length of the shortest path the gecko can take is \boxed{4\sqrt{17}} feet.
Therefore, the length of the shortest path the gecko can take to reach the fly is $\boxed{4\sqrt{17}}$ feet.
</think>