Show that the vectors $3 \hat{i} - 2 \hat{j} + \hat{k}$, $\hat{i} - 3 \hat{j} + 5 \hat{k}$ and $2 \hat{i} + \hat{j} - 4 \hat{k}$ form a right angled triangle.
Let $\overrightarrow{a} = 3 \hat{i} - 2 \hat{j} + \hat{k}$, $\;$ $\overrightarrow{b} = \hat{i} - 3 \hat{j} + 5 \hat{k}$, $\;$ $\overrightarrow{c} = 2 \hat{i} + \hat{j} - 4 \hat{k}$
$\left|\overrightarrow{a}\right| = \sqrt{\left(3\right)^2 + \left(-2\right)^2 + \left(1\right)^2} = \sqrt{14}$
$\left|\overrightarrow{b}\right| = \sqrt{\left(1\right)^2 + \left(-3\right)^2 + \left(5\right)^2} = \sqrt{35}$
$\left|\overrightarrow{c}\right| = \sqrt{\left(2\right)^2 + \left(1\right)^2 + \left(-4\right)^2} = \sqrt{21}$
Now, $\;$ $\left|\overrightarrow{a}\right|^2 + \left|\overrightarrow{c}\right|^2 = 14 + 21 = 35 = \left|\overrightarrow{b}\right|^2$ $\;\;\; \cdots \; (1)$
and $\;$ $\overrightarrow{a} \cdot \overrightarrow{c} = \left(3 \hat{i} - 2 \hat{j} + \hat{k}\right) \cdot \left(2 \hat{i} + \hat{j} - 4 \hat{k}\right) = 6 - 2 - 4 = 0$
$\implies$ $\overrightarrow{a} \perp \overrightarrow{c}$ $\;\;\; \cdots \; (2)$
$\therefore$ $\;$ From equations $(1)$ and $(2)$, the vectors $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ form a right angled triangle.