If $\bar{a}, \bar{b}, \bar{c}$ are three non-zero vectors,no two of them are collinear,$\bar{a}+2 \bar{b}$ is collinear with $\bar{c}$,and $\bar{b}+3 \bar{c}$ is collinear with $\bar{a}$,then $\bar{a}+2 \bar{b}$ is equal to:

If three points $A$,$B$,and $C$ have position vectors $(1, x, 3)$,$(3, 4, 7)$,and $(y, -2, -5)$ respectively and if they are collinear,then $(x, y)$ is

If $O$ is the circumcentre and $O'$ is the orthocentre of the triangle $ABC$,then $\overrightarrow{O'A} + \overrightarrow{O'B} + \overrightarrow{O'C} = $

$a, b, c$ are non-coplanar vectors. If $a+3 b+4 c=x(a-2 b+3 c)+y(a+5 b-2 c)+z(6 a+14 b+4 c)$,then $x+y+z=$

Let the three sides of a triangle $ABC$ be represented by the vectors $\vec{AB} = 2\hat{i}-\hat{j}+\hat{k}$,$\vec{BC} = 3\hat{i}-4\hat{j}-4\hat{k}$,and $\vec{CA} = \hat{i}-3\hat{j}-5\hat{k}$. Let $G$ be the centroid of the triangle $ABC$. Then $6(|\overrightarrow{AG}|^2+|\overrightarrow{BG}|^2+|\overrightarrow{CG}|^2)$ is equal to

If $2 \hat{i}-\hat{j}+\hat{k}$ and $\hat{i}-3 \hat{j}-5 \hat{k}$ are the position vectors of the points $A$ and $B$ respectively,$C$ divides $AB$ in the ratio $2:3$ and $M$ is the mid-point of $AB$,then $5(\text{position vector of } C) - 2(\text{position vector of } M) =$