Explanation

(a)
Let $\overrightarrow{\mathbf{OA}} = \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}}\ and\ \overrightarrow{\mathbf{OB}} = 2\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} + 4\widehat{\mathbf{k}}$
Let point $C(x_{1},y_{1},z_{1})$ divide $AB$ in the ratio 1:2
$\therefore\ x_{1} = \frac{2 + 2}{1 + 2} = \frac{4}{3},\ \ y_{1} = \frac{- 1 + 4}{1 + 2} = \frac{3}{3} = 1$
$and\ z_{1} = \frac{4 + 6}{1 + 2} = \frac{10}{3}$
Again let point $D\left( x_{2},y_{2},z_{2} \right)$ divides $AB$ in the ratio 2:1, then
$x_{2} = \frac{4 + 1}{2 + 1} = \frac{5}{3},\ \ \ y_{2} = \frac{- 2 + 2}{2 + 1} = 0$
$and\ z_{2} = \frac{8 + 3}{2 + 1} = \frac{11}{3}$
So, position vector of the points of trisection of $AB$ are position vector of
$C = - \frac{4}{3}\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + \frac{10}{3}\widehat{\mathbf{k}}$
and position vector of
$D = \frac{5}{3}\widehat{\dot{\mathbf{i}}} + \frac{11}{3}\widehat{\mathbf{k}}$
(Diagram)