Explanation

(d) Position vector of the point at which force is acting
$\overrightarrow{r_{1}} = \widehat{i} + 2\widehat{j} + 3\widehat{k}$
But we have to calculate the torque about another point. So its position vector about that another point
$\overrightarrow{r_{1}^{'}} = \overrightarrow{r_{1}} - \overrightarrow{r_{2}} = \left( \widehat{i} + 2\widehat{j} + 3\widehat{k} \right) - (3\widehat{i} - 2\widehat{j} - 3\widehat{k})$
$= - 2\widehat{i} + 4\widehat{j} + 6\widehat{k}$
Now, $\overrightarrow{\tau} = \overrightarrow{r_{1}^{'}} \times \overrightarrow{F} = ( - 2\widehat{i} + 4\widehat{j} + 6\widehat{k}) \times (4\widehat{i} - 5\widehat{j} + 3\widehat{k})$
$\overrightarrow{\tau} = \left| \begin{matrix}
\widehat{i} & \widehat{j} & \widehat{k} \\
- 2 & 4 & 6 \\
4 & - 5 & 3
\end{matrix} \right| = \widehat{i}(12 + 30) - \widehat{j}( - 6 - 24) + \widehat{k}(10 - 16)$
$= (42\widehat{i} + 30\widehat{j} - 6\widehat{k})N - m$