Find the area of the parallelogram $ABCD$ whose vertices are $A \left(-5, 2, 5\right)$, $B \left(-3, 6, 7\right)$, $C \left(4, -1, 5\right)$ and $D \left(2, -5, 3\right)$.
Given: $\;$ Vertices $\;$ $A \left(-5, 2, 5\right)$, $B \left(-3, 6, 7\right)$, $C \left(4, -1, 5\right)$, $D \left(2, -5, 3\right)$
position vector of $A$ $= - 5 \hat{i} + 2 \hat{j} + 5 \hat{k}$
position vector of $B$ $= - 3 \hat{i} + 6 \hat{j} + 7 \hat{k}$
position vector of $C$ $= 4 \hat{i} - \hat{j} + 5 \hat{k}$
position vector of $D$ $= 2 \hat{i} - 5 \hat{j} + 3 \hat{k}$
$\begin{aligned}
\text{Diagonal } \overrightarrow{AC} = \overrightarrow{d_1} & = \text{position vector of C} - \text{position vector of A} \\\\
& = \left(4 \hat{i} - \hat{j} + 5 \hat{k}\right) - \left(- 5 \hat{i} + 2 \hat{j} + 5 \hat{k}\right) \\\\
& = 9 \hat{i} - 3 \hat{j}
\end{aligned}$
$\begin{aligned}
\text{Diagonal } \overrightarrow{BD} = \overrightarrow{d_2} & = \text{position vector of D} - \text{position vector of B} \\\\
& = \left(2 \hat{i} - 5 \hat{j} + 3 \hat{k}\right) - \left(- 3 \hat{i} + 6 \hat{j} + 7 \hat{k}\right) \\\\
& = 5 \hat{i} - 11 \hat{j} - 4 \hat{k}
\end{aligned}$
Area of parallelogram $= \dfrac{1}{2} \left|\overrightarrow{d_1} \times \overrightarrow{d_2}\right|$
$\begin{aligned}
\overrightarrow{d_1} \times \overrightarrow{d_2} & = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
9 & - 3 & 0 \\
5 & -11 & -4
\end{vmatrix} \\\\
& = \hat{i} \left(12\right) - \hat{j} \left(-36\right) + \hat{k} \left(-99 + 15\right) \\\\
& = 12 \hat{i} + 36 \hat{j} - 84 \hat{k} \\\\
& = 12 \left(\hat{i} + 3 \hat{j} - 7 \hat{k}\right)
\end{aligned}$
$\therefore$ $\;$ $\left|\overrightarrow{d_1} \times \overrightarrow{d_2}\right| = 12 \times \sqrt{\left(1\right)^2 + \left(3\right)^2 + \left(-7\right)^2} = 12 \sqrt{59}$
$\therefore$ $\;$ Area of parallelogram $= \dfrac{1}{2} \times 12 \times \sqrt{59} = 6 \sqrt{59}$ sq units