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

If the position vector of a point $A$ is $a + 2b$ and a point $P$ divides $AB$ in the ratio $2:3$,where the position vector of $P$ is $a$,then the position vector of $B$ is:

If $G$ is the centroid of the $\triangle ABC$,then $\vec{GA} + \vec{GB} + \vec{GC}$ is equal to

Find a vector of magnitude $11$ in the direction opposite to that of $\overrightarrow{PQ}$,where $P$ and $Q$ are the points $(1,3,2)$ and $(-1,0,8)$ respectively.

Let $\vec{\alpha} = (\lambda - 2) \vec{a} + \vec{b}$ and $\vec{\beta} = (4\lambda - 2)\vec{a} + 3\vec{b}$ be two given vectors where $\vec{a}$ and $\vec{b}$ are non-collinear. The value of $\lambda$ for which vectors $\vec{\alpha}$ and $\vec{\beta}$ are collinear is:

If $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}+2 \hat{k},$ find the unit vector in the direction of
$(i)$ $6 \vec{b}$
(ii) $2 \vec{a}-\vec{b}$