If the points whose position vectors are $2 \hat{i}+\hat{j}+\hat{k}$, $6 \hat{i}-\hat{j}+2 \hat{k}$, and $14 \hat{i}-5 \hat{j}+p \hat{k}$ are collinear, then the value of $p$ is:

If $\hat{i}+4 \hat{j}+3 \hat{k}$,$\hat{i}+2 \hat{j}+3 \hat{k}$,and $3 \hat{i}+2 \hat{j}+\hat{k}$ are position vectors of $A$,$B$,and $C$ respectively,and if $D$ and $E$ are midpoints of sides $BC$ and $AC$,then $\overrightarrow{DE}$ is equal to:

Let $\overrightarrow{OA}=\hat{i}+2 \hat{j}-2 \hat{k}$ and $\overrightarrow{OB}=-2 \hat{i}-3 \hat{j}+6 \hat{k}$ be the position vectors of two points $A$ and $B$. If $C$ is a point on the bisector of $\angle AOB$ and $OC=\sqrt{42}$,then $\overrightarrow{OC}=$

If $\bar{a}, \bar{b}, \bar{c}$ are three unit vectors such that $|\bar{a}+\bar{b}|^2+|\bar{a}+\bar{c}|^2=8$,then $|\bar{a}+3\bar{b}|^2+|\bar{a}+3\bar{c}|^2=$

Find the centroid of the triangle whose vertices are $i + 2j, 2i + j, i + j + k$.

Let $O$ be the origin, $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$. If $R$ is a point on $\vec{OP}$ such that $\vec{OP} = 5\vec{OR}$, and $M$ is a point such that $\vec{OQ} = 5\vec{RM}$, then $\vec{PM}$ is equal to: