Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $(2 \vec{a}+\vec{b})$ and $(\vec{a}-3 \vec{b})$ externally in the ratio $1:2$. Also,show that $P$ is the mid-point of the line segment $RQ$.

(N/A) Given position vectors are $\overrightarrow{OP} = 2\vec{a} + \vec{b}$ and $\overrightarrow{OQ} = \vec{a} - 3\vec{b}$.
Point $R$ divides the line segment $PQ$ externally in the ratio $m:n = 1:2$. The section formula for external division is $\overrightarrow{OR} = \frac{m\overrightarrow{OQ} - n\overrightarrow{OP}}{m-n}$.
Substituting the values,we get:
$\overrightarrow{OR} = \frac{1(\vec{a} - 3\vec{b}) - 2(2\vec{a} + \vec{b})}{1 - 2}$
$= \frac{\vec{a} - 3\vec{b} - 4\vec{a} - 2\vec{b}}{-1}$
$= \frac{-3\vec{a} - 5\vec{b}}{-1} = 3\vec{a} + 5\vec{b}$.
To show $P$ is the mid-point of $RQ$,we calculate the mid-point of $RQ$:
Mid-point $= \frac{\overrightarrow{OR} + \overrightarrow{OQ}}{2} = \frac{(3\vec{a} + 5\vec{b}) + (\vec{a} - 3\vec{b})}{2}$
$= \frac{4\vec{a} + 2\vec{b}}{2} = 2\vec{a} + \vec{b} = \overrightarrow{OP}$.
Since the mid-point of $RQ$ is $P$,$P$ is the mid-point of the line segment $RQ$.