Show that the four points $A, \; B, \; C, \; D$ whose position vectors are $6 \hat{i} - 7 \hat{j}$, $\;$ $16 \hat{i} - 19 \hat{j} - 4 \hat{k}$, $\;$ $3 \hat{j} - 6 \hat{k}$ and $2 \hat{i} - 5 \hat{j} + 10 \hat{k}$ respectively are coplanar.
Given: $\;$ Position vector (p.v) of point $A = 6 \hat{i} - 7 \hat{j}$
p.v of point $B = 16 \hat{i} - 19 \hat{j} - 4 \hat{k}$
p.v of point $C = 3 \hat{j} - 6 \hat{k}$
p.v of point $D = 2 \hat{i} - 5 \hat{j} + 10 \hat{k}$
Now,
$\begin{aligned}
\overrightarrow{AB} & = \text{p.v of point B} - \text{p.v of point A} \\\\
& = \left(16 \hat{i} - 19 \hat{j} - 4 \hat{k}\right) - \left(6 \hat{i} - 7 \hat{j}\right) \\\\
& = 10 \hat{i} - 12 \hat{j} - 4 \hat{k}
\end{aligned}$
$\begin{aligned}
\overrightarrow{AC} & = \text{p.v of point C} - \text{p.v of point A} \\\\
& = \left(3 \hat{j} - 6 \hat{k}\right) - \left(6 \hat{i} - 7 \hat{j}\right) \\\\
& = -6 \hat{i} + 10 \hat{j} - 6 \hat{k}
\end{aligned}$
$\begin{aligned}
\overrightarrow{AD} & = \text{p.v of point D} - \text{p.v of point A} \\\\
& = \left(2 \hat{i} - 5 \hat{j} + 10 \hat{k}\right) - \left(6 \hat{i} - 7 \hat{j}\right) \\\\
& = -4 \hat{i} + 2 \hat{j} + 10 \hat{k}
\end{aligned}$
Scalar triple product of the vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$ and $\overrightarrow{AD}$ is
$\begin{aligned}
\left[\overrightarrow{AB} \;\; \overrightarrow{AC} \;\; \overrightarrow{AD}\right] & = \begin{vmatrix}
10 & - 12 & -4 \\
- 6 & 10 & -6 \\
-4 & 2 & 10
\end{vmatrix} \\\\
& = 10 \left(100 + 12\right) + 12 \left(- 60 - 24\right) - 4 \left(- 12 + 40\right) \\\\
& = 1120 - 1008 - 112 \\\\
& = 0
\end{aligned}$
$\because \;$ $\left[\overrightarrow{AB} \;\; \overrightarrow{AC} \;\; \overrightarrow{AD}\right] = 0$ $\implies$ Vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$ and $\overrightarrow{AD}$ are coplanar.
$\implies$ Points $A, \; B, \; C$ and $D$ are coplanar.