The vertices of a triangle have position vectors $4 \hat{i} + 5 \hat{j} + 6 \hat{k}$, $5 \hat{i} + 6 \hat{j} + 4 \hat{k}$, $6 \hat{i} + 4 \hat{j} + 5 \hat{k}$. Prove that the triangle is equilateral.
Let $A$, $B$ and $C$ be the vertices of the triangle.
Let
position vector of $A = 4 \hat{i} + 5 \hat{j} + 6 \hat{k}$
position vector of $B = 5 \hat{i} + 6 \hat{j} + 4 \hat{k}$
position vector of $C = 6 \hat{i} + 4 \hat{j} + 5 \hat{k}$
Now,
$\begin{aligned}
\overrightarrow{AB} & = \text{position vector of B} - \text{position vector of A} \\\\
& = \left(5 \hat{i} + 6 \hat{j} + 4 \hat{k}\right) - \left(4 \hat{i} + 5 \hat{j} + 6 \hat{k}\right) \\\\
& = \hat{i} + \hat{j} - 2 \hat{k}
\end{aligned}$
$\begin{aligned}
\overrightarrow{BC} & = \text{position vector of C} - \text{position vector of B} \\\\
& = \left(6 \hat{i} + 4 \hat{j} + 5 \hat{k}\right) - \left(5 \hat{i} + 6 \hat{j} + 4 \hat{k}\right) \\\\
& = \hat{i} - 2 \hat{j} + \hat{k}
\end{aligned}$
$\begin{aligned}
\overrightarrow{CA} & = \text{position vector of A} - \text{position vector of C} \\\\
& = \left(4 \hat{i} + 5 \hat{j} + 6 \hat{k}\right) - \left(6 \hat{i} + 4 \hat{j} + 5 \hat{k}\right) \\\\
& = -2 \hat{i} + \hat{j} + \hat{k}
\end{aligned}$
$\left|\overrightarrow{AB}\right| = \sqrt{\left(1\right)^2 + \left(1\right)^2 + \left(-2\right)^2} = \sqrt{6}$
$\left|\overrightarrow{BC}\right| = \sqrt{\left(1\right)^2 + \left(-2\right)^2 + \left(1\right)^2} = \sqrt{6}$
$\left|\overrightarrow{CA}\right| = \sqrt{\left(-2\right)^2 + \left(1\right)^2 + \left(1\right)^2} = \sqrt{6}$
$\therefore$ $\;$ $\left|\overrightarrow{AB}\right| = \left|\overrightarrow{BC}\right| = \left|\overrightarrow{CA}\right|$
$\implies$ $A$, $B$ and $C$ are the vertices of an equilateral triangle.