If $r \cdot i = r \cdot j = r \cdot k$ and $|r| = 3$, then $r = $

$ABCD$ is a tetrahedron. $\bar{i}-2\bar{j}+3\bar{k}$, $-2\bar{i}+\bar{j}+3\bar{k}$, and $3\bar{i}+2\bar{j}-\bar{k}$ are the position vectors of the points $A, B, C$ respectively. $-\bar{i}+2\bar{j}-3\bar{k}$ is the position vector of the centroid of the triangular face $BCD$. If $G$ is the centroid of the tetrahedron, then $GD=$

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

Let $A$ and $B$ be points with position vectors $a$ and $b$ with respect to the origin $O$. If the point $C$ on $OA$ is such that $2AC = CO$,$CD$ is parallel to $OB$ and $|\overrightarrow{CD}| = 3|\overrightarrow{OB}|$,then $\overrightarrow{AD}$ is equal to

If $\vec{p} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{q} = \hat{i} + \hat{j} - \hat{k}$,and $\vec{a}$ and $\vec{b}$ are two vectors such that $\vec{p} = 2\vec{a} + \vec{b}$ and $\vec{q} = \vec{a} + 2\vec{b}$,then the angle between $\vec{a}$ and $\vec{b}$ is:

If $\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}$,then