If $\vec a,\vec b, \vec c$ are three non-zero vectors which are pairwise non-collinear. If $\vec a +3\vec b$ is collinear with $\vec c$ and $\vec b+2\vec c$ is collinear with $\vec a$, then $\vec a +3\vec b +6\vec c$, is:

$\vec 0$

It is given that $\vec a +3\vec b$ is collinear with $\vec c$ and $\vec b+2\vec a$ is collinear with $\vec a$. Therefore,

$\vec a+3\vec b=λ\vec c$ and $\vec b+2\vec c=μ\vec a$ for some $λ, μ$.

$⇒\vec a+3(μ\vec a-2\vec c)=λ\vec c$  [On eliminating $\vec b$]

$⇒(1+3μ)\vec a-(λ+6)\vec c=\vec 0$

$⇒1+3μ=0$ and $λ+6=0$  [∵ $\vec a$ and $\vec c$ are non-collinear]

$⇒λ=-6$ and $μ=-\frac{1}{3}$

$∴\vec a+3\vec b=λ\vec c⇒\vec a+3\vec b+6\vec c=\vec 0$