Let $\overrightarrow{a},\overrightarrow{b}$and $\overrightarrow{c}$ be three non–zero vectors such that no two of these are collinear. If the vector $\overrightarrow{a}+2\overrightarrow{b}$ is collinear with $\overrightarrow{c}$ and $\overrightarrow{b}+3\overrightarrow{c}$ is collinear with $\overrightarrow{a}$( $\lambda$ being some non–zero scalar), then $\overrightarrow{a}+2\overrightarrow{b}+6\overrightarrow{c}$ equals 

given $\overrightarrow{a}+2\overrightarrow{b}$.....(i)

and  $\overrightarrow{b}+3\overrightarrow{c}=\lambda \overrightarrow{a}\Rightarrow \overrightarrow{b}=\lambda \overrightarrow{a}-3\overrightarrow{c}$ On putting this value in (i), 

we get $\overrightarrow{a}+2(\lambda \overrightarrow{a}-3\overrightarrow{c})=t\overrightarrow{c}\Rightarrow \overrightarrow{a}+2\lambda \overrightarrow{a}-6\overrightarrow{c}=t\overrightarrow{c}$

$\Rightarrow (\overrightarrow{a}-6\overrightarrow{c})=t\overrightarrow{c}-2\lambda \overrightarrow{a}$

 $\overrightarrow{a}$ and $\overrightarrow{c}$ are non-collinear

$\therefore 1=-2\lambda \Rightarrow \lambda =-1/2\text{ and }-6=t\Rightarrow t=-6$

From (i), we get $\overrightarrow{a}+2\overrightarrow{b}=-6\overrightarrow{c}\Rightarrow \overrightarrow{a}+2\overrightarrow{b}+6\overrightarrow{c}=\overrightarrow{0}$