If $a$ and $b$ are the position vectors of $A$ and $B$ respectively,then the position vector of a point $C$ on $AB$ produced such that $\overrightarrow{AC} = 3\overrightarrow{AB}$ is

Let $O$ be the origin,$A$ and $B$ be two points with position vectors $-3 \hat{i}-3 \hat{j}+4 \hat{k}$ and $4 \hat{i}-4 \hat{j}-3 \hat{k}$ respectively. Let $P$ be a point such that the line drawn through $P$ parallel to $\overrightarrow{OB}$ meets $OA$ in $L$ and another line through $P$ parallel to $\overrightarrow{OA}$ meets $OB$ in $M$. If $L$ divides $OA$ in the ratio $2:3$ and $M$ divides $OB$ in the ratio $3:2$,then the distance from $O$ to $P$ is

Let $p = (x + 4y)\vec{a} + (2x + y + 1)\vec{b}$ and $q = (y - 2x + 2)\vec{a} + (2x - 3y - 1)\vec{b}$,where $\vec{a}$ and $\vec{b}$ are non-collinear vectors. If $3p = 2q$,then the values of $x$ and $y$ are:

If three points $A$,$B$,and $C$ have position vectors $(1, x, 3)$,$(3, 4, 7)$,and $(y, -2, -5)$ respectively and if they are collinear,then $(x, y)$ is

If $\vec{r} = 3\hat{i} + 2\hat{j} - 5\hat{k}$,$\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + 3\hat{j} - 2\hat{k}$,and $\vec{c} = -2\hat{i} + \hat{j} - 3\hat{k}$ such that $\vec{r} = \lambda\vec{a} + \mu\vec{b} + \gamma\vec{c}$,then -

In a triangle $ABC$, if $\overline{BC} = \bar{i} - 2\bar{j} + 2\bar{k}$ and $\overline{CA} = 6\bar{i} + 3\bar{j} - 2\bar{k}$, then the perimeter of the triangle is