Let $A$ and $B$ be two points. The position vector of $A$ is $6b - 2a$. Point $P$ divides the line segment $AB$ in the ratio $1 : 2$. If $a - b$ is the position vector of $P$,what is the position vector of $B$?

If $\hat{a}, \hat{b}$ and $\hat{c}$ are non-coplanar vectors and if $\hat{d}$ is such that $\hat{d} = \frac{1}{x}(\hat{a} + \hat{b} + \hat{c})$ and $\hat{d} = \frac{1}{y}(\hat{b} + \hat{c} + \hat{d})$ where $x$ and $y$ are non-zero real numbers,then $\frac{1}{xy}(\hat{a} + \hat{b} + \hat{c} + \hat{d})$ equals to

If $a=2 \hat{i}+3 \hat{j}+\hat{k}$,$b=\hat{i}-3 \hat{j}-5 \hat{k}$ and $c=3 \hat{i}-4 \hat{k}$,then match the items of List-$I$ with those of List-$II$.

$A$. Unit vector in the direction opposite to that $a-b$ is	$(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$
$B$. If $\vec{AB} = a, \vec{BC} = b$,then $\vec{CA} =$	$(ii) \ 2 \hat{i} - \frac{8}{3} \hat{k}$
$C$. If $a, b, c$ are the position vectors of the vertices of a triangle then,its centroid is	$(iii) \ -3 \hat{i} + 4 \hat{k}$
$D$. If $d$ is a vector of magnitude $2 \sqrt{14}$ and parallel to the vector $a$,then $b + d =$	$(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$
	$(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$

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

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be any three non-coplanar vectors. If $m$ and $n$ are scalars such that $\vec{a}+\vec{b}=m \vec{d}-\vec{c}$ and $\vec{b}+\vec{c}=n \vec{a}-\vec{d}$,then $3 \vec{a}+2 \vec{b}+2 \vec{c}+\vec{d}=$