The points with position vectors $\bar{a}+\bar{b}$,$\bar{a}-\bar{b}$,and $\bar{a}+k\bar{b}$ are collinear:

The position vectors of the points $A$ and $B$ are $\vec{a}$ and $\vec{b}$ respectively. If the position vector of the point $C$ is $\frac{\vec{a}}{2} + \frac{\vec{b}}{3}$, then:

Find the position vector of the midpoint of the vector joining the points $P(2, 3, 4)$ and $Q(4, 1, -2)$.

If $\hat{i}-2 \hat{j}+3 \hat{k}$,$2 \hat{i}+3 \hat{j}+\hat{k}$,and $-3 \hat{i}-\hat{j}-2 \hat{k}$ are the position vectors of three points $A$,$B$,and $C$ respectively,then $A$,$B$,and $C$

Five points given by $A, B, C, D, E$ are in a plane. Three forces $\overrightarrow{AC}, \overrightarrow{AD},$ and $\overrightarrow{AE}$ act at $A$,and three forces $\overrightarrow{CB}, \overrightarrow{DB},$ and $\overrightarrow{EB}$ act at $B$. Then their resultant is:

Find the position vector of a point $R$ which divides the line joining the two points $P$ and $Q$ with position vectors $\overrightarrow{OP} = 2\vec{a} + \vec{b}$ and $\overrightarrow{OQ} = \vec{a} - 2\vec{b}$ respectively,in the ratio $1:2$,$(i)$ internally and (ii) externally.