If $a = (1, -1)$ and $b = (-2, m)$ are two collinear vectors,then $m = $

If $\vec{a}, \vec{b}, \vec{c}$ are three vectors such that $|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$,then $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$ does not exceed:

$A, B, C, D, E$ are five coplanar points,then $\overrightarrow{DA} + \overrightarrow{DB} + \overrightarrow{DC} + \overrightarrow{AE} + \overrightarrow{BE} + \overrightarrow{CE}$ is equal to

Three non-zero non-collinear vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are such that $\vec{a}+3\vec{b}$ is collinear with $\vec{c}$, and $3\vec{b}+2\vec{c}$ is collinear with $\vec{a}$. Then $\vec{a}+3\vec{b}+2\vec{c}$ equals to

$ABCD$ is a parallelogram such that $L$ is the mid-point of $BC$. Then,$\vec{AL}$ is equal to:

If $\vec{a}$ and $\vec{b}$ are the position vectors $(P.V.)$ of two points $A$ and $B$ respectively,and $C$ divides the line segment $AB$ in the ratio $2 : 1$,then the position vector of $C$ is: