The figure formed by the four points $(\hat{i}+\hat{j}-\hat{k}), (2\hat{i}+3\hat{j}), (5\hat{j}-2\hat{k})$ and $(\hat{k}-\hat{j})$ is

The position vectors of the points $A, B, C$ are $(2i + j - k)$,$(3i - 2j + k)$,and $(i + 4j - 3k)$ respectively. These points

Let $A, B$ and $C$ be three points on a circle of radius $R$. If $O$ is the centre of the circle and $\angle AOB = 45^{\circ}, \angle BOC = 45^{\circ}$,then the resultant of $\vec{OA}, \vec{OB}$ and $\vec{OC}$ has magnitude

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

Find a unit vector in the direction of $\overrightarrow{PQ},$ where $P$ and $Q$ have coordinates $(5, 0, 8)$ and $(3, 3, 2),$ respectively.

The vectors $a = 2\hat{i} + 3\hat{j} + 6\hat{k}$ and $b$ are collinear and $|b| = 21$,then $b =$