If $\vec{PO} + \vec{OQ} = \vec{QO} + \vec{OR}$,then

$A$ vector of magnitude $\sqrt{2}$ units along the internal bisector of the angle between the vectors $\vec{a} = 2 \hat{i} - 2 \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ is

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

If $\overrightarrow{AB} = (2\hat{i} + 3\hat{j} - \hat{k})$ and $A(1, 2, -1)$ is the given point,then the coordinates of $B$ are

The vectors $\vec{AB} = 3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$ are the sides of a $\triangle ABC$. The length of the median through $A$ is:

$ABCD$ 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 $GD=$