Show that the points $A$, $B$, $C$ with position vectors $\;$ $-2 \overrightarrow{a} + 3 \overrightarrow{b} + 5 \overrightarrow{c}$, $\;$ $\overrightarrow{a} + 2 \overrightarrow{b} + 3 \overrightarrow{c}$ $\;$ and $\;$ $7 \overrightarrow{a}- \overrightarrow{c}$ $\;$ respectively, are collinear.
Let $O$ be the origin.
The position vectors of points $A$, $B$ and $C$ are $\;$ $-2 \overrightarrow{a} + 3 \overrightarrow{b} + 5 \overrightarrow{c}$, $\;$ $\overrightarrow{a} + 2 \overrightarrow{b} + 3 \overrightarrow{c}$ $\;$ and $\;$ $7 \overrightarrow{a} - \overrightarrow{c}$ $\;$ respectively.
i.e. $\;$ $\overrightarrow{OA} = -2 \overrightarrow{a} + 3 \overrightarrow{b} + 5 \overrightarrow{c}$
$\overrightarrow{OB} = \overrightarrow{a} + 2 \overrightarrow{b} + 3 \overrightarrow{c}$ $\;$ and
$\overrightarrow{OC} = 7 \overrightarrow{a} - \overrightarrow{c}$
Now, $\;$ $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$
i.e. $\;$ $\overrightarrow{AB} = \left(\overrightarrow{a} + 2 \overrightarrow{b} + 3 \overrightarrow{c}\right) - \left(- 2 \overrightarrow{a} + 3 \overrightarrow{b} + 5 \overrightarrow{c}\right) = 3 \overrightarrow{a} - \overrightarrow{b} - 2 \overrightarrow{c}$
and $\;$ $\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB}$
i.e. $\;$ $\overrightarrow{BC} = \left(7 \overrightarrow{a} - \overrightarrow{c}\right) - \left(\overrightarrow{a} + 2 \overrightarrow{b} + 3 \overrightarrow{c}\right) = 6 \overrightarrow{a} - 2 \overrightarrow{b} - 4 \overrightarrow{c}$
i.e. $\;$ $\overrightarrow{BC} = 2 \left(3 \overrightarrow{a} - \overrightarrow{b} - 2 \overrightarrow{c}\right) = 2 \overrightarrow{AB}$
$\implies$ $\overrightarrow{AB}$ $\;$ and $\;$ $\overrightarrow{BC}$ $\;$ are parallel vectors and point $B$ is common to both of them.
$\implies$ $\overrightarrow{AB}$ $\;$ and $\;$ $\overrightarrow{BC}$ $\;$ are collinear vectors.
$\implies$ $A$, $B$ and $C$ are collinear points.