Find the area of a triangle having the points $A(1, 1, 1)$, $B(1, 2, 3)$ and $C(2, 3, 1)$ as its vertices.

$\frac{\sqrt{21}}{2}$

The correct answer is Option (2) → $\frac{\sqrt{21}}{2}$ ##

We have $\vec{AB} = \hat{j} + 2\hat{k}$ and $\vec{AC} = \hat{i} + 2\hat{j}$. The area of the given triangle is $\frac{1}{2}|\vec{AB} \times \vec{AC}|$.

Now, $\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix} = -4\hat{i} + 2\hat{j} - \hat{k}$

Therefore $|\vec{AB} \times \vec{AC}| = \sqrt{16+4+1} = \sqrt{21}$

Thus, the required area is $\frac{1}{2}\sqrt{21}$