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

Explanation

To find the area of the triangle with vertices A(1, -1, 2), B(2, 1, -1), and C(3, -1, 2), we can use the formula for the area of a triangle formed by three points in 3D space. The area can be calculated using the cross product of two vectors formed by these points.

### Step 1: Find the vectors AB and AC
First, we need to find the vectors AB and AC.

- **Vector AB**:
\[ \vec{AB} = \vec{B} - \vec{A} = (2 - 1, 1 - (-1), -1 - 2) = (1, 2, -3) \]

- **Vector AC**:
\[ \vec{AC} = \vec{C} - \vec{A} = (3 - 1, -1 - (-1), 2 - 2) = (2, 0, 0) \]

### Step 2: Calculate the cross product of vectors AB and AC
Next, we calculate the cross product \(\vec{AB} \times \vec{AC}\).

\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 0 & 0 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} 2 & -3 \\ 0 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -3 \\ 2 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ 2 & 0 \end{vmatrix} \]

Calculating each of the determinants:

1. \(\hat{i}(2 \cdot 0 - (-3) \cdot 0) = 0\)
2. \(-\hat{j}(1 \cdot 0 - (-3) \cdot 2) = -(-6) = 6\)
3. \(\hat{k}(1 \cdot 0 - 2 \cdot 2) = -4\)

Putting it all together:

\[ \vec{AB} \times \vec{AC} = 0\hat{i} + 6\hat{j} - 4\hat{k} = (0, 6, -4) \]

### Step 3: Find the magnitude of the cross product
Now, we find the magnitude of the cross product:

\[ |\vec{AB} \times \vec{AC}| = \sqrt{0^2 + 6^2 + (-4)^2} = \sqrt{0 + 36 + 16} = \sqrt{52} = 2\sqrt{13} \]

### Step 4: Calculate the area of the triangle
The area \(A\) of the triangle is given by:

\[ A = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \cdot 2\sqrt{13} = \sqrt{13} \]

Thus, the area of the triangle is \(\sqrt{13}\).

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