Show that the points whose position vectors given by $-2 \hat{i} + 3 \hat{j} + 5 \hat{k}$, $\hat{i} + 2 \hat{j} + 3 \hat{k}$, $7 \hat{i} - \hat{k}$ are collinear.
Let the points be $A$, $B$ and $C$.
Let $O$ be the origin.
Then, $\overrightarrow{OA} = -2 \hat{i} + 3 \hat{j} + 5 \hat{k}$; $\overrightarrow{OB} = \hat{i} + 2 \hat{j} + 5 \hat{k}$ and $\overrightarrow{OC} = 7 \hat{i} - \hat{k}$
$\begin{aligned}
\text{Now, } \overrightarrow{AB} & = \overrightarrow{OB} - \overrightarrow{OA} \\\\
& = \left(\hat{i} + 2 \hat{j} + 3 \hat{k}\right) - \left(-2 \hat{i} + 3 \hat{j} + 5 \hat{k}\right) \\\\
& = 3 \hat{i} - \hat{j} - 2 \hat{k}
\end{aligned}$
$\begin{aligned}
\text{and } \overrightarrow{AC} & = \overrightarrow{OC} - \overrightarrow{OA} \\\\
& = \left(7 \hat{i} - \hat{k}\right) - \left(-2 \hat{i} + 3 \hat{j} + 5 \hat{k}\right) \\\\
& = 9 \hat{i} - 3 \hat{j} - 6 \hat{k} \\\\
& = 3 \left(3 \hat{i} - \hat{j} - 2 \hat{k}\right) \\\\
& = 3 \overrightarrow{AB}
\end{aligned}$
$\implies$ Vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel.
Also, point $A$ is a common point to both the vectors.
$\implies$ The points $A$, $B$ and $C$ are collinear.