Let $\overline{\mathrm{a}}, \overline{\mathrm{b}}$, and $\overline{\mathrm{c}}$ be three non-zero vectors…
Let $\overline{\mathrm{a}}, \overline{\mathrm{b}}$, and $\overline{\mathrm{c}}$ be three non-zero vectors such that no two of these are collinear. If the vector $\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}$ is collinear with $\overline{\mathrm{c}}$ and $\overline{\mathrm{b}}+3 \overline{\mathrm{c}}$ is collinear with $\overline{\mathrm{a}}$, then $\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}+6 \overline{\mathrm{c}}$ equals
$\quad \lambda \bar{c}$ ( $\lambda$ being some non-zero scalar)
$\lambda \bar{b}$ ( $\lambda$ being some non-zero scalar)
$\lambda \bar{a}$ ( $\lambda$ being some non-zero scalar)
$\overline{0}$ ( $\lambda$ being some non-zero scalar)
Solution
$\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}$ is collinear with $\overline{\mathrm{c}}$
$\therefore \quad \bar{a}+2 \bar{b}=n \bar{c}$...(i)
Similarly $\overline{\mathrm{b}}+3 \overline{\mathrm{c}}=\mathrm{m} \overline{\mathrm{a}}$...(ii)
$m$ and $n$ are non-zero scalars.
$\therefore \quad$ (i) $\Rightarrow \overline{\mathrm{a}}+2 \overline{\mathrm{~b}}+6 \overline{\mathrm{c}}=(\mathrm{n}+6) \overline{\mathrm{c}}$
(ii) $\Rightarrow \bar{a}+2 \bar{b}+6 \bar{c}=(2 m+1) \bar{a}$
$\Rightarrow \mathrm{n}+6=0$ and $2 \mathrm{~m}+1=0$
$\Rightarrow \mathrm{n}=-6$ and $\mathrm{m}=\frac{-1}{2}$
$\therefore \quad$ (i) $\Rightarrow \bar{a}+2 \bar{b}+6 \bar{c}=0$