If $P, Q, R$ are the mid-points of the sides $AB, BC$ and $CA$ of $\triangle ABC$ respectively,then $PC - BQ =$

$A$ point $C$ with position vector $\frac{3 \bar{a}+4 \bar{b}-5 \bar{c}}{3}$ (where $\bar{a}, \bar{b}$ and $\bar{c}$ are non-coplanar vectors) divides the line segment joining $A$ and $B$ in the ratio $2:1$. If the position vector of $A$ is $\bar{a}-2 \bar{b}+3 \bar{c}$,then find the position vector of $B$.

If the collinear points $A, B$ and $C$ have position vectors respectively $(1, x, 3), (3, 4, 7)$ and $(y, -2, -5)$, then $x+y=$

If the position vectors of the vertices $A, B, C, D$ of a quadrilateral are $7 \hat{i}-4 \hat{j}+7 \hat{k}, \hat{i}-6 \hat{j}+10 \hat{k}, -\hat{i}-3 \hat{j}+4 \hat{k}$,and $5 \hat{i}-\hat{j}+5 \hat{k}$ respectively,then $ABCD$ is

The position vector of a point at a distance of $3\sqrt{11}$ units from $i - j + 2k$ on a line passing through the points $i - j + 2k$ and $3i + j + k$ is

If $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are position vectors of $4$ points such that $2 \vec{a}+3 \vec{b}+5 \vec{c}-10 \vec{d}=\vec{0}$,then the ratio in which the line joining $\vec{c}$ and $\vec{d}$ divides the line segment joining $\vec{a}$ and $\vec{b}$ is