Let a → , b → and c → be three non-zero vectors such that b → and c → are non-collinear if a → + 5 b → is…

Given: $\overrightarrow{\mathrm{a}}+5\overrightarrow{\mathrm{b}}$ is collinear with $\overrightarrow{\mathrm{c}}, \overrightarrow{\mathrm{b}}+6\overrightarrow{\mathrm{c}}$ is collinear with $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{a}}+\alpha \overrightarrow{\mathrm{b}}+\beta \overrightarrow{\mathrm{c}}=\overrightarrow{0}$.

$\Rightarrow \overrightarrow{\mathrm{a}}+5\overrightarrow{\mathrm{b}}=\mathrm{\lambda }\overrightarrow{\mathrm{c}} ...\left(\mathrm{i}\right)$

$\Rightarrow \overrightarrow{\mathrm{b}}+6\overrightarrow{\mathrm{c}}=\mathrm{\mu }\overrightarrow{\mathrm{a}} ...\left(\mathrm{ii}\right)$

$\Rightarrow \overrightarrow{\mathrm{a}}=\mathrm{\lambda }\overrightarrow{\mathrm{c}}-5\overrightarrow{\mathrm{b}}=\frac{\overrightarrow{\mathrm{b}}+6\overrightarrow{\mathrm{c}}}{\mathrm{\mu }}$

$\Rightarrow \mathrm{\lambda }\overrightarrow{\mathrm{c}}-5\overrightarrow{\mathrm{b}}=\frac{6}{\mathrm{\mu }}\overrightarrow{\mathrm{c}}+\frac{1}{\mathrm{\mu }}\overrightarrow{\mathrm{b}}$

Comparing both the sides,

$\Rightarrow \mathrm{\mu }=\frac{-1}{5}, \mathrm{\lambda }=-30$

Putting these values in equation $\left(\mathrm{i}\right)$

$\Rightarrow \overrightarrow{\mathrm{a}}+5\overrightarrow{\mathrm{b}}=-30\overrightarrow{\mathrm{c}}$

$\Rightarrow \overrightarrow{\mathrm{a}}+5\overrightarrow{\mathrm{b}}+30\overrightarrow{\mathrm{c}}=0$

Comparing this equation with $\overrightarrow{\mathrm{a}}+\mathrm{\alpha }\overrightarrow{\mathrm{b}}+\mathrm{\beta }\overrightarrow{\mathrm{c}}=\overrightarrow{0}$.

$\Rightarrow \mathrm{\alpha }=5, \mathrm{\beta }=30$

$\Rightarrow \mathrm{\alpha }+\mathrm{\beta }=35$