The area of the triangle whose vertices are A ( 1,-1,2) , B ( 1,2, -1) ,C ( 3, -1, 2) is ________.

Explanation

To find the area of the triangle with vertices A(1, -1, 2), B(1, 2, -1), and C(3, -1, 2), we can follow these steps:

### Step 1: Represent the vertices as vectors
We represent the points A, B, and C in vector form:
- \( \vec{A} = 1\hat{i} - 1\hat{j} + 2\hat{k} \)
- \( \vec{B} = 1\hat{i} + 2\hat{j} - 1\hat{k} \)
- \( \vec{C} = 3\hat{i} - 1\hat{j} + 2\hat{k} \)

### Step 2: Find the vectors AB and BC
Now, we find the vectors \( \vec{AB} \) and \( \vec{BC} \):
- \( \vec{AB} = \vec{B} - \vec{A} = (1 - 1)\hat{i} + (2 - (-1))\hat{j} + (-1 - 2)\hat{k} = 0\hat{i} + 3\hat{j} - 3\hat{k} = 3\hat{j} - 3\hat{k} \)
- \( \vec{BC} = \vec{C} - \vec{B} = (3 - 1)\hat{i} + (-1 - 2)\hat{j} + (2 - (-1))\hat{k} = 2\hat{i} - 3\hat{j} + 3\hat{k} \)

### Step 3: Calculate the cross product \( \vec{AB} \times \vec{BC} \)
We calculate the cross product \( \vec{AB} \times \vec{BC} \):
\[ \vec{AB} \times \vec{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & -3 \\ 2 & -3 & 3 \end{vmatrix} \]
Calculating the determinant:
- For \( \hat{i} \): \( 3 \cdot 3 - (-3)(-3) = 9 - 9 = 0 \)
- For \( \hat{j} \): \( -(0 \cdot 3 - (-3)(2)) = -(0 + 6) = -6 \)
- For \( \hat{k} \): \( 0 \cdot (-3) - 3 \cdot 2 = 0 - 6 = -6 \)

Thus, we have:
\[ \vec{AB} \times \vec{BC} = 0\hat{i} - 6\hat{j} - 6\hat{k} \]

### Step 4: Find the magnitude of the cross product
Now we find the magnitude of the cross product:
\[ |\vec{AB} \times \vec{BC}| = \sqrt{0^2 + (-6)^2 + (-6)^2} = \sqrt{0 + 36 + 36} = \sqrt{72} = 6\sqrt{2} \]

### Step 5: Calculate the area of the triangle
The area \( A \) of triangle ABC is given by:
\[ A = \frac{1}{2} |\vec{AB} \times \vec{BC}| = \frac{1}{2} (6\sqrt{2}) = 3\sqrt{2} \]

Thus, the area of the triangle is \( 3\sqrt{2} \) square units.

### Final Answer:
The area of the triangle ABC is \( 3\sqrt{2} \).

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