Let $\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}$ are  three non-zero vectors such that any two of them are non-collinear. If $\overrightarrow{\mathrm{a}}$ +2$\overrightarrow{\mathrm{b}}$  is collinear with $\overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{b}}$+3$\overrightarrow{\mathrm{c}}$ is collinear  with $\overrightarrow{\mathrm{a}}$ Then $\overrightarrow{\mathrm{a}}$+2$\overrightarrow{\mathrm{b}}$+6$\overrightarrow{\mathrm{c}}$=

Given that $\overrightarrow{\mathrm{a}}+2\overrightarrow{\mathrm{b}}$ is collinear with $\overrightarrow{\mathrm{c}}$
$\Rightarrow \overrightarrow{\mathrm{a}}+2\overrightarrow{\mathrm{b}}=\mathrm{\lambda }\overrightarrow{\mathrm{c}}$for some scalar $\mathrm{\lambda }\to 1$
 And $\overrightarrow{\mathrm{b}}+3\overrightarrow{\mathrm{c}}$ is collinear with $\overrightarrow{\mathrm{a}}$
$\Rightarrow \overrightarrow{\mathrm{b}}+3\overrightarrow{\mathrm{c}}=\mathrm{\mu }\overrightarrow{\mathrm{a}}$for some scalar $\mathrm{\mu }\to 2$
From eq1 we have $\overrightarrow{\mathrm{a}}=\mathrm{\lambda }\overrightarrow{\mathrm{c}}-2\overrightarrow{\mathrm{b}}$
Substituting the value of $\overrightarrow{\mathrm{a}}$ in 2 we get 
$\Rightarrow$$\overrightarrow{\mathrm{b}}+3\overrightarrow{\mathrm{c}}=\mathrm{\mu }(\mathrm{\lambda }\overrightarrow{\mathrm{c}}-2\overrightarrow{\mathrm{b})}$
$\Rightarrow (1+2\mathrm{\mu })\overrightarrow{\mathrm{b}}+(3-\mathrm{\mu \lambda })\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{o}}$
$\Rightarrow 1+2\mathrm{\mu }=0$ and $3-\mathrm{\mu \lambda }=0$ ($\because \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ are Non collinear vectors)
$\Rightarrow \mathrm{\mu }=-\frac{-1}{2} \mathrm{and} \mathrm{\lambda }=-6$ and $\mathrm{\lambda }=-6$
Substituting  the values of $\mathrm{\lambda }$ and $\mathrm{\mu }$ in 1 and 2 respectively 
We get $\overrightarrow{\mathrm{a}}+2\overrightarrow{\mathrm{b}}+6\overrightarrow{\mathrm{c}}=0$