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

Solution :
Let $$\vec a + 2\vec b = t\vec c$$   and $$\vec b + 3\vec c = s\vec a,$$   where $$t$$ and $$s$$ are scalars. Adding, we get
$$\eqalign{ & \vec a + 3\vec b + 3\vec c = t\vec c + s\vec a \cr & \Rightarrow \vec a + 2\vec b + 6\vec c = t\vec c + s\vec a - \vec b + 3\vec c \cr & \Rightarrow \vec a + 2\vec b + 6\vec c = t\vec c + \left( {\vec b + 3\vec c} \right) - \vec b + 3\vec c \cr & \Rightarrow \vec a + 2\vec b + 6\vec c = \left( {t + 6} \right)\vec c\,\,\,\,\,\,\,\left[ {{\text{using }}s\vec a = \vec b + 3\vec c} \right] \cr & \Rightarrow \vec a + 2\vec b + 6\vec c = \lambda \vec c,{\text{ where }}\lambda = t + 6 \cr} $$