Let $D, E, F$ be the midpoints of the sides $BC, CA, AB$ of a triangle $ABC$ respectively. Then $\vec{AD} + \vec{BE} + \vec{CF} = \dots$

Given $p = 2a - 3b$,$q = a - 2b + c$,and $r = -3a + b + 2c$,where $a, b,$ and $c$ are non-zero,non-coplanar vectors,then the vector $-2a + 3b - c$ is equal to:

If $M_1, M_2, M_3$ and $M_4$ are respectively the magnitudes of the vectors $\vec{a}_1 = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{a}_2 = -3\hat{i} - 4\hat{j} - 4\hat{k}$,$\vec{a}_3 = -\hat{i} + \hat{j} - \hat{k}$,and $\vec{a}_4 = -\hat{i} + 3\hat{j} + \hat{k}$,then the correct order of $M_1, M_2, M_3$ and $M_4$ is:

The points with position vectors $60\,i + 3\,j$,$40\,i - 8\,j$,and $a\,i - 52\,j$ are collinear,if $a = $

$a, b, c$ 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 $x+y+z=$

If the vectors $\vec{AB} = \hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} + \hat{j} + 2\hat{k}$ are two sides of a triangle $ABC$,whose centroid is $G$,then $|\vec{AG}| = $