<Question>
The points $O,$ $A,$ and $B$ satisfy $\overrightarrow{OA} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} -3 \\ -2 \\ 1 \end{pmatrix}.$  Find the area of triangle $OAB.$
</Question>

<think>
<Plan_1>
Okay, so I need to find the area of triangle OAB given the vectors OA and OB. Let me recall how to do this. Hmm, I think the area of a triangle formed by two vectors can be found using the cross product. The formula is (1/2) times the magnitude of the cross product of the two vectors. Right? So, if I have vectors OA and OB, then the area is (1/2)|OA × OB|. Let me make sure that's correct.

First, let me visualize the problem.

<prompt_1.1> Visualize the problem using the given vectors OA and OB. </prompt_1.1>

Wait, but cross product gives a vector perpendicular to both OA and OB, and its magnitude is equal to the area of the parallelogram formed by the vectors. So, half of that would be the area of the triangle.
Yes, that's right. So, the plan is to compute the cross product of OA and OB, then find its magnitude, and divide by 2.

Alright, let's write down the vectors:

<prompt_1.2> Write down the given vectors OA and OB. </prompt_1.2>
-----

Alternatively, another way to compute the area is using the formula: |OA × OB| / 2. Which is the same as the cross product method. So, both ways lead to the same result.
Therefore, the answer should be 7√3.
-----

Alternatively, maybe I can verify using coordinates. Let's see.

<prompt_1.3> Consider using coordinates to verify the area of the triangle. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
Points O, A, and B are in space, I assume, since the vectors are in 3D. The triangle is formed by these three points. Since O is the origin, A is at (1,2,3), and B is at (-3,-2,1). To find the area of triangle OAB, I can use the cross product method because that's the standard way to find the area when you have vectors from the origin.
</execution_1.1>

<execution_1.2>
Vector OA is \(\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\) and vector OB is \(\begin{pmatrix} -3 \\ -2 \\ 1 \end{pmatrix}\).
</execution_1.2>

<execution_1.3>
If O is the origin, then the coordinates of A are (1,2,3) and coordinates of B are (-3,-2,1). Then, the area of triangle OAB can be calculated using the formula for the area with coordinates. But in 3D, it's a bit more involved. However, the cross product method is the standard approach here. Alternatively, we can use the formula involving vectors. But since we already did the cross product, I think that's sufficient.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.2:
First, compute the cross product OA × OB. The cross product of two vectors \(\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}\) and \(\begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}\) is given by the determinant of a matrix with the standard unit vectors i, j, k in the first row, the components of the first vector in the second row, and the components of the second vector in the third row.

So, let's compute that determinant.

<prompt_2.1> Calculate the cross product OA × OB using the determinant method. </prompt_2.1>
-----

But let me just check if there's another method, maybe using vectors and the magnitude formula. Let's see.

Alternatively, the area can be calculated using Heron's formula, but for that, I need the lengths of all three sides.

<prompt_2.2> Explore the possibility of using vectors and the magnitude formula to verify the area. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
That is:

\[
\overrightarrow{OA} \times \overrightarrow{OB} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & 3 \\
-3 & -2 & 1 \\
\end{vmatrix}
\]

Calculating this determinant, we get:

\mathbf{i} \cdot (2 \cdot 1 - 3 \cdot (-2)) - \mathbf{j} \cdot (1 \cdot 1 - 3 \cdot (-3)) + \mathbf{k} \cdot (1 \cdot (-2) - 2 \cdot (-3))
\]

Let me compute each component step by step.

For the \(\mathbf{i}\) component: (2)(1) - (3)(-2) = 2 + 6 = 8.

For the \(\mathbf{j}\) component: Note there's a negative sign in front. So, -(1*1 - 3*(-3)) = -(1 + 9) = -10.

For the \(\mathbf{k}\) component: (1)(-2) - (2)(-3) = -2 + 6 = 4.

So, putting it all together, the cross product is:

\[
\begin{pmatrix} 8 \\ -10 \\ 4 \end{pmatrix}
\]
</execution_2.1>

<execution_2.2>
Let me see if that's feasible here.

First, compute the lengths of OA, OB, and AB.

Length of OA is sqrt(1² + 2² + 3²) = sqrt(1 + 4 + 9) = sqrt(14).

Length of OB is sqrt((-3)² + (-2)² + 1²) = sqrt(9 + 4 + 1) = sqrt(14).

Length of AB is the distance between points A and B. Coordinates of A are (1,2,3), coordinates of B are (-3,-2,1). So, AB vector is OB - OA = (-3 -1, -2 -2, 1 -3) = (-4, -4, -2). So, the length of AB is sqrt((-4)^2 + (-4)^2 + (-2)^2) = sqrt(16 + 16 + 4) = sqrt(36) = 6.

So, sides are OA = sqrt(14), OB = sqrt(14), AB = 6. Then, using Heron's formula, the semi-perimeter s = (sqrt(14) + sqrt(14) + 6)/2 = (2*sqrt(14) + 6)/2 = sqrt(14) + 3.

Then, the area is sqrt[s(s - OA)(s - OB)(s - AB)].
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Now, I need to find the magnitude of this vector.
The magnitude is calculated as:

<prompt_3.1> Calculate the magnitude of the cross product vector. </prompt_3.1>
-----

Based on execution_2.2:
Let me compute that.

<prompt_3.2> Calculate the area using Heron's formula as described in the above plan. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
\[
| \overrightarrow{OA} \times \overrightarrow{OB} | = \sqrt{8^2 + (-10)^2 + 4^2} = \sqrt{64 + 100 + 16} = \sqrt{180}
\]

Simplify sqrt(180). Let's see, 180 = 36 * 5, so sqrt(36*5) = 6*sqrt(5). Therefore, the magnitude is 6√5.
</execution_3.1>

<execution_3.2>
s = sqrt(14) + 3

s - OA = sqrt(14) + 3 - sqrt(14) = 3

s - OB = same as above, 3

s - AB = sqrt(14) + 3 - 6 = sqrt(14) - 3

So, the area is sqrt[ (sqrt(14) + 3)(3)(3)(sqrt(14) - 3) ]

Simplify the expression inside the square root:

First, note that (sqrt(14) + 3)(sqrt(14) - 3) = (sqrt(14))² - (3)^2 = 14 - 9 = 5

Therefore, the product becomes 5 * 3 * 3 = 5*9=45

Thus, the area is sqrt(45) = 3*sqrt(5). Wait, but that contradicts the previous result of 7*sqrt(3). That can't be right. There must be a mistake here.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Wait, but let me double-check my cross product calculation because if that's wrong, the rest is wrong.

<prompt_4.1> Verify the cross product calculation for potential errors. </prompt_4.1>
-----

Based on execution_3.2:
Wait, hold on. Let me check Heron's formula again.

<prompt_4.2> Re-evaluate the Heron's formula calculation to identify any mistakes. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
Let me recalculate the cross product step by step.

First, the determinant:

i*(2*1 - 3*(-2)) - j*(1*1 - 3*(-3)) + k*(1*(-2) - 2*(-3)).

Calculating each component:

i: 2*1 = 2; 3*(-2) = -6; so 2 - (-6) = 8. So, i*8.

j: 1*1 = 1; 3*(-3) = -9; so 1 - (-9) = 10. But since it's minus j, it becomes -10j.

k: 1*(-2) = -2; 2*(-3) = -6; so -2 - (-6) = -2 + 6 = 4. So, k*4.

Yes, that seems correct. So, cross product is (8, -10, 4). Then, magnitude squared is 64 + 100 + 16 = 180. Square root of 180 is indeed 6*sqrt(5). Then, half of that is 3*sqrt(5). So, area is 3√5.

But wait, this contradicts the cross product method? Wait, no, hold on. Wait, no, the cross product method gives (1/2)|OA × OB|, which is (1/2)(6√5) = 3√5. But earlier, when I thought of 7√3, that was a miscalculation. Wait, no, let me go back.
</execution_4.1>

<execution_4.2>
Heron's formula is sqrt[s(s - a)(s - b)(s - c)]. So, in this case:

s = (sqrt(14) + sqrt(14) + 6)/2 = sqrt(14) + 3

Then,

s - a = sqrt(14) + 3 - sqrt(14) = 3

s - b = same as above, 3

s - c = sqrt(14) + 3 - 6 = sqrt(14) - 3

Therefore, the product is [sqrt(14) + 3][3][3][sqrt(14) - 3] = 9[sqrt(14) + 3][sqrt(14) - 3] = 9[(14) - 9] = 9[5] = 45. So, sqrt(45) = 3*sqrt(5). But this contradicts the cross product result of 7*sqrt(3).
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Wait, no. Wait, cross product gives |OA × OB| = 6√5, so area is (1/2)(6√5) = 3√5. So, why did I think earlier that I had 7√3? Maybe I confused the magnitude with something else. Wait, let me check again.

<prompt_5.1> Correct the area calculation using the cross product method. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
Wait, no. Wait, let's retrace.

Wait, the cross product is (8, -10, 4). Then, the magnitude is sqrt(8² + (-10)^2 + 4^2) = sqrt(64 + 100 + 16) = sqrt(180). sqrt(180) is sqrt(36*5) = 6*sqrt(5). Then, half of that is 3*sqrt(5). So, area is 3√5. So, that's the answer. Then, why did I initially think it was 7√3? Maybe that was a mistake. Wait, let's check the cross product again.

Wait, maybe I miscalculated the cross product. Let me check each component again.

For the i component: (2)(1) - (3)(-2) = 2 + 6 = 8. Correct.

For the j component: It's (1)(1) - (3)(-3) = 1 + 9 = 10. But since it's the j component in the determinant, which is subtracted, so -10. Correct.

For the k component: (1)(-2) - (2)(-3) = -2 + 6 = 4. Correct. So, cross product is (8, -10, 4). Then, magnitude is sqrt(8² + 10² + 4²) = sqrt(64 + 100 + 16) = sqrt(180) = 6√5. Then, area is 1/2 * 6√5 = 3√5. So, 3√5 is the answer.

But wait, maybe the problem is in 3D, so Heron's formula should also work, but when I tried Heron's formula, I also got 3√5. Wait, but in my initial calculation, I thought I had 7√3, but that must have been an error. Let me check again.

Wait, in the Heron's formula, the sides are OA = sqrt(14), OB = sqrt(14), AB = 6. So, semi-perimeter s = (sqrt(14) + sqrt(14) + 6)/2 = sqrt(14) + 3. Then, s - OA = 3, s - OB = 3, s - AB = sqrt(14) - 3. Then, the product is [sqrt(14) + 3][3][3][sqrt(14) - 3] = 9[(sqrt(14))^2 - 9] = 9[14 - 9] = 9*5 = 45. So, sqrt(45) = 3√5. So, same as cross product. Therefore, the area is 3√5. So, that's the answer.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Wait, but in the initial problem statement, the vectors OA and OB are given as OA = (1,2,3) and OB = (-3,-2,1). Then, when I computed the cross product, I got (8, -10, 4), which has magnitude sqrt(64 + 100 + 16) = sqrt(180) = 6√5. Then, half of that is 3√5. So, that's the area. So, why did I think of 7√3? Maybe because sqrt(14) is approximately 3.74, and sqrt(14) + sqrt(14) + 6 is about 3.74*2 +6 = 13.48, so semi-perimeter ~6.74, and then the area would be sqrt(6.74*(6.74 - 3.74)*(6.74 - 3.74)*(6.74 -6)) which is sqrt(6.74*3*3*0.74). Let me compute that:

6.74 * 3 * 3 * 0.74 ≈ 6.74 * 9 * 0.74 ≈ 6.74 * 6.66 ≈ 44.9, which is approximately 45, so sqrt(44.9) ≈ 6.7, but that's not 3√5, since 3√5 ≈ 3*2.236 ≈ 6.708. So, that's correct. Therefore, 3√5 is the exact value.

Wait, so why did I think of 7√3 earlier? Maybe because I miscalculated the cross product? Let me check again.

<prompt_6.1> Verify the cross product calculation once more to ensure accuracy. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
Wait, cross product:

i component: (2)(1) - (3)(-2) = 2 +6=8. Correct.

j component: -[(1)(1) - (3)(-3)] = -[1 +9] = -10. Correct.

k component: (1)(-2) - (2)(-3) = -2 +6=4. Correct. So, cross product is (8, -10, 4). Then, magnitude squared is 8² + (-10)^2 +4²=64+100+16=180. sqrt(180)=6√5. So, area is 3√5. So, that's correct. So, the answer is 3√5. So, why did I think of 7√3? Maybe because sqrt(14) is approx 3.74, and 3.74*3=11.22, but that's not relevant. Wait, perhaps I confused the cross product components. Wait, let me check once again. Wait, cross product is (8, -10, 4). Then, magnitude is sqrt(8^2 + (-10)^2 +4^2)=sqrt(64+100+16)=sqrt(180)=6√5. Yes. Then, area is 1/2 of that, so 3√5. So, that's correct. Therefore, the answer is 3√5. So, why did I initially think 7√3? Maybe because sqrt(14) is involved in the lengths, but no. So, the correct answer is 3√5.
</execution_6.1>
</Execution_6>

<Final_answer>
Therefore, the area of triangle $OAB$ is $\boxed{3\sqrt{5}}$.
</Final_answer>
</think>