Find the area of the triangle whose vertices are $\left(3, -1, 2\right)$, $\left(1, -1, -3\right)$ and $\left(4, -3, 1\right)$
Let $A \left(3, -1, 2\right)$, $B \left(1, -1, -3\right)$ and $C \left(4, -3, 1\right)$
Position vector (p.v) of point $A = 3 \hat{i} - \hat{j} + 2 \hat{k}$
p.v of point $B = \hat{i} - \hat{j} - 3 \hat{k}$
p.v of point $C = 4 \hat{i} - 3 \hat{j} + \hat{k}$
$\begin{aligned}
\overrightarrow{AB} & = \text{p.v of point B - p.v of point A} \\\\
& = \left(\hat{i} - \hat{j} - 3 \hat{k}\right) - \left(3 \hat{i} - \hat{j} + 2 \hat{k}\right) \\\\
& = -2 \hat{i} - 5 \hat{k}
\end{aligned}$
$\begin{aligned}
\overrightarrow{AC} & = \text{p.v of point C - p.v of point A} \\\\
& = \left(4 \hat{i} - 3 \hat{j} + \hat{k}\right) - \left(3 \hat{i} - \hat{j} + 2 \hat{k}\right) \\\\
& = \hat{i} - 2 \hat{j} - \hat{k}
\end{aligned}$
Area of $\triangle ABC = \dfrac{1}{2} \left|\overrightarrow{AB} \times \overrightarrow{AC} \right|$
$\begin{aligned}
\overrightarrow{AB} \times \overrightarrow{AC} & = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-2 & 0 & -5 \\
1 & -2 & -1
\end{vmatrix} \\\\
& = \hat{i} \left(0 - 10\right) - \hat{j} \left(2 + 5\right) + \hat{k} \left(4 - 0\right) \\\\
& = - 10 \hat{i} - 7 \hat{j} + 4 \hat{k}
\end{aligned}$
$\therefore$ $\;$ $\left|\overrightarrow{AB} \times \overrightarrow{AC}\right| = \sqrt{\left(-10\right)^2 + \left(-7\right)^2 + \left(4\right)^2} = \sqrt{165}$
$\therefore$ $\;$ Area of $\triangle ABC = \dfrac{1}{2} \sqrt{165}$ sq units