Let `A(1, 2, 3), B(0, 0, 1) and C(-1, 1, 1)` are the vertices of `triangleABC`.
Q. The area of`(triangleABC)` is equal to

Options

A. `(9)/(2)`
B. `(sqrt(17))/(2)`
C. `(17)/(2)`
D. `(7)/(2)`

Answer

To find the area of triangle ABC with vertices A(1, 2, 3), B(0, 0, 1), and C(-1, 1, 1), we will use the formula for the area of a triangle in three-dimensional space. The area can be calculated using the formula:

\[ \text{Area} = \frac{1}{2} \sqrt{|\vec{AB} \times \vec{AC}|^2} \]

Where \(\vec{AB}\) and \(\vec{AC}\) are the vectors from A to B and A to C, respectively.

### Step 1: Find the vectors \(\vec{AB}\) and \(\vec{AC}\)

\[ \vec{AB} = B - A = (0 - 1, 0 - 2, 1 - 3) = (-1, -2, -2) \]
\[ \vec{AC} = C - A = (-1 - 1, 1 - 2, 1 - 3) = (-2, -1, -2) \]

### Step 2: Calculate the cross product \(\vec{AB} \times \vec{AC}\)

Using the determinant formula for the cross product:

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

Calculating the determinant:

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

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} -2 & -2 \\ -1 & -2 \end{vmatrix} = (-2)(-2) - (-2)(-1) = 4 - 2 = 2\)

2. \(\begin{vmatrix} -1 & -2 \\ -2 & -2 \end{vmatrix} = (-1)(-2) - (-2)(-2) = 2 - 4 = -2\)

3. \(\begin{vmatrix} -1 & -2 \\ -2 & -1 \end{vmatrix} = (-1)(-1) - (-2)(-2) = 1 - 4 = -3\)

Putting it all together:

\[ \vec{AB} \times \vec{AC} = \hat{i}(2) - \hat{j}(-2) + \hat{k}(-3) = 2\hat{i} + 2\hat{j} - 3\hat{k} \]

### Step 3: Find the magnitude of the cross product

\[ |\vec{AB} \times \vec{AC}| = \sqrt{(2)^2 + (2)^2 + (-3)^2} = \sqrt{4 + 4 + 9} = \sqrt{17} \]

### Step 4: Calculate the area of triangle ABC

\[ \text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \sqrt{17} \]

Thus, the area of triangle ABC is:

\[ \text{Area} = \frac{\sqrt{17}}{2} \]

### Final Answer
The area of triangle ABC is \(\frac{\sqrt{17}}{2}\).

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