The vertices of a `DeltaABC` are A(2, 3, 5), B(-1, 3, 2) and C (3,5,-2). The area of `DeltaABC` is

Options

A. `8sqrt3`
B. `6sqrt2`
C. `8sqrt2`
D. `9sqrt2`

Answer

To find the area of triangle ABC with vertices A(2, 3, 5), B(-1, 3, 2), and C(3, 5, -2), we can follow these steps:

### Step 1: Find the vectors AB and AC
To find the vectors AB and AC, we subtract the coordinates of the points.

- **AB vector**:
\[ \text{AB} = B - A = (-1 - 2, 3 - 3, 2 - 5) = (-3, 0, -3) \]

- **AC vector**:
\[ \text{AC} = C - A = (3 - 2, 5 - 3, -2 - 5) = (1, 2, -7) \]

### Step 2: Compute the cross product of vectors AB and AC
To find the area of the triangle, we need to compute the cross product \( \text{AB} \times \text{AC} \).

Using the determinant method:
\[ \text{AB} \times \text{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 0 & -3 \\ 1 & 2 & -7 \end{vmatrix} \]

Calculating the determinant:
\[ = \mathbf{i} \begin{vmatrix} 0 & -3 \\ 2 & -7 \end{vmatrix} - \mathbf{j} \begin{vmatrix} -3 & -3 \\ 1 & -7 \end{vmatrix} + \mathbf{k} \begin{vmatrix} -3 & 0 \\ 1 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \( \mathbf{i} \):
\[ 0 \cdot (-7) - (-3) \cdot 2 = 0 + 6 = 6 \]
2. For \( \mathbf{j} \):
\[ -3 \cdot (-7) - (-3) \cdot 1 = 21 + 3 = 24 \]
3. For \( \mathbf{k} \):
\[ -3 \cdot 2 - 0 \cdot 1 = -6 \]

Putting it all together:
\[ \text{AB} \times \text{AC} = (6, -24, -6) \]

### Step 3: Find the magnitude of the cross product
The magnitude of the vector \( \text{AB} \times \text{AC} \) is given by:
\[ \left| \text{AB} \times \text{AC} \right| = \sqrt{6^2 + (-24)^2 + (-6)^2} \]
Calculating:
\[ = \sqrt{36 + 576 + 36} = \sqrt{648} = \sqrt{36 \cdot 18} = 6\sqrt{18} \]

### Step 4: Calculate the area of triangle ABC
The area of triangle ABC is given by:
\[ \text{Area} = \frac{1}{2} \left| \text{AB} \times \text{AC} \right| = \frac{1}{2} \cdot 6\sqrt{18} = 3\sqrt{18} \]
We can simplify \( \sqrt{18} \) as \( \sqrt{9 \cdot 2} = 3\sqrt{2} \):
\[ \text{Area} = 3 \cdot 3\sqrt{2} = 9\sqrt{2} \]

### Final Answer
The area of triangle ABC is \( 9\sqrt{2} \).

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