NIMCET 2016 Question Paper with solutions

Given that a, b and c are non-zero vectors and $\mathit{a}+2\mathit{b}$ is collinear with c and $\mathit{b}+3\mathit{c}$ is collinear with a.
$\therefore \mathit{a}+2\mathit{b}=\mathit{\lambda }\mathit{c}$....(i)
and $\mathit{b}+3\mathit{c}=\mathit{\mu }\mathit{a}$....(ii)  [from same λ,µ]
Put the value of b in Eq. (i) by Eq (ii), we get
$\mathit{a}+2\left(\mathit{\mu }\mathit{a}-3\mathit{c}\right)=\mathit{\lambda }\mathit{c}$
$\Rightarrow \mathit{a}+2\mathit{\mu }\mathit{a}-6\mathit{c}=\mathit{\lambda }\mathit{c}$
$\Rightarrow \mathit{a}\left(1+2\mathit{\mu }\right)-\mathit{c}\left(\mathit{\lambda }+6\right)=0$  [∵ a and c are non-collinear]
$\Rightarrow 1+2\mathit{\mu }=0$ and $\mathit{\lambda }+6=0$
$\Rightarrow \mathit{\lambda }=-6$
and $\mathit{\mu }=-\frac{1}{2}$
$\because \mathit{a}+2\mathit{b}=\mathit{\lambda }\mathit{c}$
$\therefore \mathit{a}+2\mathit{b}=-6\mathit{c}$
$\Rightarrow \mathit{a}+2\mathit{b}+6\mathit{c}=0$