If $\bar{a}, \bar{b}, \bar{c}$ are three unit vectors such that $|\bar{a}+\bar{b}|^2+|\bar{a}+\bar{c}|^2=8$,then $|\bar{a}+3\bar{b}|^2+|\bar{a}+3\bar{c}|^2=$

Find the position vector of the point which divides the join of the points with position vectors $2\vec{a} - 3\vec{b}$ and $3\vec{a} - 2\vec{b}$ externally in the ratio $2:3$.

If $2\vec{a} - 3\vec{b}$,$\vec{b}$,and $\vec{a} - \vec{b}$ are the position vectors of three points $A$,$B$,and $C$ respectively,then they are:

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

If the vectors $\overline{AB}=3 \hat{i}+4 \hat{k}$ and $\overline{AC}=5 \hat{i}-2 \hat{j}+4 \hat{k}$ are the sides of the triangle $ABC$,then the length of the median through $A$ is:

Given $p = 2a - 3b$,$q = a - 2b + c$,and $r = -3a + b + 2c$,where $a, b,$ and $c$ are non-zero,non-coplanar vectors,then the vector $-2a + 3b - c$ is equal to: