Okay, so I need to find the volume enclosed by the set S where the sum of the squares of the distances from any point P to three given planes is equal to 36. Let me start by understanding the problem step by step.

First, the set S is defined by the equation d₁² + d₂² + d₃² = 36, where d₁, d₂, d₃ are the distances from a point P to three planes. Each distance is calculated using the formula for the distance from a point to a plane. Since the equation involves the squares of these distances, this might resemble the equation of a sphere, but in three-dimensional space. However, the planes here are not necessarily orthogonal, so the resulting set S might not be a sphere but some kind of quadratic surface. Alternatively, it might be a sphere if the planes are orthogonal, but I need to check that first.

Let me recall the formula for the distance from a point (x, y, z) to a plane ax + by + cz + d = 0. The distance is |ax + by + cz + d| / sqrt(a² + b² + c²). So, I need to write each of the given planes in the standard form and compute their distances accordingly.

The three planes given are:
1. x - z = 0. This can be written as x + 0y - z + 0 = 0. So, a = 1, b = 0, c = -1, d = 0.
2. x - 2y + z = 0. This is already in standard form: x - 2y + z + 0 = 0. So, a = 1, b = -2, c = 1, d = 0.
3. x + y + z = 0. Also in standard form: x + y + z + 0 = 0. So, a = 1, b = 1, c = 1, d = 0.

Therefore, the distances from a point P(x, y, z) to each plane are:

d₁ = |x - z| / sqrt(1² + 0² + (-1)²) = |x - z| / sqrt(2)
d₂ = |x - 2y + z| / sqrt(1² + (-2)² + 1²) = |x - 2y + z| / sqrt(6)
d₃ = |x + y + z| / sqrt(1² + 1² + 1²) = |x + y + z| / sqrt(3)

Then, the equation defining S is:

(d₁)² + (d₂)² + (d₃)² = 36

Substituting the expressions for d₁, d₂, d₃:

(|x - z| / sqrt(2))² + (|x - 2y + z| / sqrt(6))² + (|x + y + z| / sqrt(3))² = 36

Since squaring removes the absolute value, this simplifies to:

(x - z)² / 2 + (x - 2y + z)² / 6 + (x + y + z)² / 3 = 36

Now, I need to simplify this equation and see what geometric shape it represents. Let's expand each term and combine like terms.

First term: (x - z)² / 2 = (x² - 2xz + z²) / 2

Second term: (x - 2y + z)² / 6. Let's expand the numerator:

(x - 2y + z)² = x² + ( -2y )² + z² + 2*(x*(-2y)) + 2*(x*z) + 2*(-2y*z) = x² + 4y² + z² - 4xy + 2xz - 4yz

So, dividing by 6: (x² + 4y² + z² - 4xy + 2xz - 4yz)/6

Third term: (x + y + z)² / 3. Expand the numerator:

(x + y + z)² = x² + y² + z² + 2xy + 2xz + 2yz

Divide by 3: (x² + y² + z² + 2xy + 2xz + 2yz)/3

Now, sum all three terms:

First term + Second term + Third term = [ (x² - 2xz + z²)/2 ] + [ (x² + 4y² + z² - 4xy + 2xz - 4yz)/6 ] + [ (x² + y² + z² + 2xy + 2xz + 2yz)/3 ] = 36

To combine these, let's find a common denominator. The denominators are 2, 6, and 3. The least common denominator is 6.

Multiply each term by 6 to eliminate denominators:

First term: 6*(x² - 2xz + z²)/2 = 3*(x² - 2xz + z²) = 3x² - 6xz + 3z²

Second term: 6*(x² + 4y² + z² - 4xy + 2xz - 4yz)/6 = x² + 4y² + z² - 4xy + 2xz - 4yz

Third term: 6*(x² + y² + z² + 2xy + 2xz + 2yz)/3 = 2*(x² + y² + z² + 2xy + 2xz + 2yz) = 2x² + 2y² + 2z² + 4xy + 4xz + 4yz

Now, sum all three results:

3x² - 6xz + 3z² + x² + 4y² + z² - 4xy + 2xz - 4yz + 2x² + 2y² + 2z² + 4xy + 4xz + 4yz

Combine like terms:

x² terms: 3x² + x² + 2x² = 6x²

y² terms: 4y² + 2y² = 6y²

z² terms: 3z² + z² + 2z² = 6z²

xy terms: -4xy + 4xy = 0

xz terms: -6xz + 2xz + 4xz = 0

yz terms: -4yz + 4yz = 0

Wait, so all cross terms cancel out? Let me check:

- For xy terms: -4xy (from first expansion) and +4xy (from the third expansion). Yes, they cancel.

- For xz terms: -6xz (from first) +2xz (from second) +4xz (from third): (-6 + 2 + 4) = 0

- For yz terms: -4yz (from second) +4yz (from third): 0

So, all cross terms cancel out. Then, the equation simplifies to:

6x² + 6y² + 6z² = 36

Divide both sides by 6:

x² + y² + z² = 6

Wait, that's the equation of a sphere with radius sqrt(6) centered at the origin. But hold on, how is this possible? The original equation was a sum of squared distances to three planes, which after simplification turns into the equation of a sphere. That seems counterintuitive. Let me verify my calculations again.

First term expansion:

(x - z)^2 / 2: (x² - 2xz + z²)/2

Second term: (x - 2y + z)^2 /6: (x² +4y² + z² -4xy +2xz -4yz)/6

Third term: (x + y + z)^2 /3: (x² + y² + z² +2xy +2xz +2yz)/3

Sum:

First term: (x² - 2xz + z²)/2

Second term: (x² +4y² + z² -4xy +2xz -4yz)/6

Third term: (x² + y² + z² +2xy +2xz +2yz)/3

Adding them together:

First term multiplied by 3: 3*(x² -2xz + z²)/2*3? Wait, no. Wait, when I multiplied each term by 6 to eliminate denominators, perhaps I made a mistake there.

Wait, let me correct that. Let's handle the addition without multiplying by 6 first.

Original terms:

Term1: (x² -2xz + z²)/2

Term2: (x² +4y² + z² -4xy +2xz -4yz)/6

Term3: (x² + y² + z² +2xy +2xz +2yz)/3

To add them, let's compute each component:

x² terms: Term1 has x²/2, Term2 has x²/6, Term3 has x²/3. So total x²: (1/2 + 1/6 + 1/3) x² = (3/6 + 1/6 + 2/6) x² = 6/6 x² = x²

Similarly, y² terms: Term1 has 0, Term2 has 4y²/6 = 2y²/3, Term3 has y²/3. Total y²: (2/3 + 1/3) y² = y²

z² terms: Term1 has z²/2, Term2 has z²/6, Term3 has z²/3. Total z²: (1/2 + 1/6 + 1/3) z² = same as x², which is z².

Now, cross terms:

For xy terms: Term2 has -4xy/6 = -2xy/3, Term3 has 2xy/3. Total: (-2xy/3 + 2xy/3) = 0

For xz terms: Term1 has -2xz/2 = -xz, Term2 has 2xz/6 = xz/3, Term3 has 2xz/3. Total: (-xz + xz/3 + 2xz/3) = (-xz + xz) = 0

For yz terms: Term2 has -4yz/6 = -2yz/3, Term3 has 2yz/3. Total: (-2yz/3 + 2yz/3) = 0

So, indeed, all cross terms cancel out. Therefore, the equation reduces to:

x² + y² + z² = 36 * (1/2 + 1/6 + 1/3) ?

Wait, no. Wait, the original sum of terms was:

Term1 + Term2 + Term3 = 36

But when adding Term1 + Term2 + Term3, we found that:

Term1 + Term2 + Term3 = x² + y² + z² = 36

Wait, that can't be. Wait, the sum of Term1 + Term2 + Term3 is equal to 36. But after simplifying, we found that Term1 + Term2 + Term3 = x² + y² + z². Therefore, x² + y² + z² = 36. So the set S is the sphere of radius sqrt(36) = 6, centered at the origin.

Wait, but that seems conflicting with the original problem statement. The original problem states that the sum of squared distances to three planes is 36. After simplifying, we get x² + y² + z² = 36. So S is a sphere of radius 6. Therefore, the volume enclosed by S is the volume of a sphere with radius 6.

Wait, but in that case, the answer would be straightforward: (4/3)πr³ = (4/3)π(6)³ = (4/3)π*216 = 288π. But that seems too easy. The problem mentions three planes, so maybe I made a mistake in the simplification.

Wait, let me check again.

Original equation:

d₁² + d₂² + d₃² = 36

Where d₁, d₂, d₃ are the distances to the three planes.

After expanding each distance squared and adding them, we get x² + y² + z² = 36. Therefore, S is the sphere of radius 6. Then the volume is (4/3)π*(6)³ = 288π. But that seems too straightforward for a problem that mentions three planes. Maybe I made a mistake in the expansion.

Wait, let me verify the expansion once again.

First term: (x - z)^2 / 2 = (x² - 2xz + z²)/2

Second term: (x - 2y + z)^2 /6 = (x² +4y² +z² -4xy +2xz -4yz)/6

Third term: (x + y + z)^2 /3 = (x² + y² + z² +2xy +2xz +2yz)/3

Adding them:

Term1: x²/2 - xz + z²/2

Term2: x²/6 + (4y²)/6 + z²/6 - (4xy)/6 + (2xz)/6 - (4yz)/6

Simplify term2: x²/6 + (2y²)/3 + z²/6 - (2xy)/3 + (xz)/3 - (2yz)/3

Term3: x²/3 + y²/3 + z²/3 + (2xy)/3 + (2xz)/3 + (2yz)/3

Now, adding Term1 + Term2 + Term3:

x² terms: (1/2 + 1/6 + 1/3) x² = (3/6 + 1/6 + 2/6) x² = 6/6 x² = x²

y² terms: (2/3 + 1/3) y² = y²

z² terms: (1/2 + 1/6 + 1/3) z² = same as x², so z²

Cross terms:

xz terms: -xz + (xz)/3 + (2xz)/3 = -xz + xz = 0

xy terms: -(2xy)/3 + (2xy)/3 = 0

yz terms: -(2yz)/3 + (2yz)/3 = 0

Therefore, indeed, the sum simplifies to x² + y² + z² = 36. Therefore, S is the sphere of radius 6. Therefore, the volume is (4/3)π*(6)^3 = 288π.

But the problem mentions three planes, so maybe there's a catch here. Let me think again. If the three planes are orthogonal, then the sum of squared distances might correspond to a sphere. But in general, for arbitrary planes, the set of points with the sum of squared distances equal to a constant is an ellipsoid. However, in this case, after simplifying, it turned out to be a sphere. That must be because the planes have certain symmetries or relationships.

Let me check if the planes are orthogonal. The three planes are:

1. x - z = 0

2. x - 2y + z = 0

3. x + y + z = 0

Compute the dot products of their normal vectors to check orthogonality.

Normal vector of first plane: (1, 0, -1)

Normal vector of second plane: (1, -2, 1)

Dot product: 1*1 + 0*(-2) + (-1)*1 = 1 + 0 -1 = 0. So first and second planes are orthogonal.

Normal vector of third plane: (1, 1, 1)

Dot product with first plane's normal: 1*1 + 1*0 + 1*(-1) = 1 + 0 -1 = 0. So first and third planes are orthogonal.

Dot product with second plane's normal: 1*1 + 1*(-2) + 1*1 = 1 -2 +1 = 0. So second and third planes are also orthogonal.

Therefore, all three planes are mutually orthogonal. Therefore, the coordinate system defined by these planes is an orthogonal coordinate system. In such a case, the equation d₁² + d₂² + d₃² = constant represents a sphere in the coordinate system defined by the three planes.

But in the standard Cartesian coordinate system, this would correspond to a sphere. However, since the planes are not the standard x, y, z planes, but rather a rotated system, the equation might still represent a sphere in the original coordinate system.

Wait, but in our calculation, we transformed the equation into the standard sphere equation in the original coordinates. Therefore, the region S is indeed a sphere of radius 6, and the volume is 288π.

But let me check again with a different approach. Suppose we consider the coordinate system defined by the three planes. Since the planes are mutually orthogonal, we can think of them as defining a new orthogonal coordinate system (like a rotated Cartesian system). In this new system, the equation d₁² + d₂² + d₃² = 36 would represent a sphere of radius 6. However, when transformed back to the original coordinate system, since the coordinate system is orthogonal, the sphere remains a sphere. Therefore, the volume enclosed by S is the same as the volume of a sphere with radius 6.

But why does the problem mention three planes? Because if they weren't orthogonal, the equation would represent an ellipsoid. But here, since the planes are orthogonal, the equation becomes a sphere. So perhaps the answer is indeed 288π.

But let me verify with a specific point. For example, take the origin (0,0,0). The distances to all three planes are zero, so d₁² + d₂² + d₃² = 0 ≠ 36. So the origin is not on S. Take a point along one of the coordinate axes. Let's say along the x-axis: (a, 0, 0). Compute distances:

d₁ = |a - 0| / sqrt(2) = |a| / sqrt(2)

d₂ = |a - 0 + 0| / sqrt(6) = |a| / sqrt(6)

d₃ = |a + 0 + 0| / sqrt(3) = |a| / sqrt(3)

Then, d₁² + d₂² + d₃² = (a² / 2) + (a² / 6) + (a² / 3) = (3a² + a² + 2a²)/6 = 6a² /6 = a²

Set equal to 36: a² = 36, so a = ±6. Therefore, the point (6, 0, 0) is on S, and so are (0,6,0) and (0,0,6). Wait, but according to the previous conclusion, the sphere has radius 6, so the point (6,0,0) is on the sphere. However, in the original coordinate system, is that point on the sphere? Wait, the sphere in the original coordinate system is x² + y² + z² = 36. So (6,0,0) is on that sphere. Similarly, (0,6,0) and (0,0,6) are also on the sphere. Therefore, this checks out.

But wait, when we calculated the distance from (6,0,0) to each plane:

d₁ = |6 - 0| / sqrt(2) = 6 / sqrt(2) = 3*sqrt(2)

d₂ = |6 - 0 + 0| / sqrt(6) = 6 / sqrt(6) = sqrt(6)

d₃ = |6 + 0 + 0| / sqrt(3) = 6 / sqrt(3) = 2*sqrt(3)

Then, d₁² + d₂² + d₃² = (18) + (6) + (12) = 36. Which matches 36. So that's correct.

Therefore, the conclusion is that S is a sphere of radius 6 centered at the origin, and the volume is 288π. Therefore, the answer is \boxed{288\pi}.

But wait, just to be thorough, let me check another point. Let's take a point off the axes, say (3,3,3). Compute distances:

d₁ = |3 - 3| / sqrt(2) = 0

d₂ = |3 - 6 + 3| / sqrt(6) = |0| / sqrt(6) = 0

d₃ = |3 + 3 + 3| / sqrt(3) = 9 / sqrt(3) = 3*sqrt(3)

Then, d₁² + d₂² + d₃² = 0 + 0 + 27 = 27 ≠ 36. So this point is not on S. Let's check if (3,3,3) is on the sphere x² + y² + z² = 36. 9 + 9 + 9 = 27 ≠ 36. So it's not on the sphere. Therefore, the set S is indeed the sphere x² + y² + z² = 36, and the volume is 288π. Therefore, the answer is correct.

**Final Answer**
\boxed{288\pi}
Thus, the region of the volume enclosed by $S$ is $\boxed{288\pi}$.
</think>