Forces $2 \hat{i} + 7 \hat{j}$, $\;$ $2 \hat{i} - 5 \hat{j} + 6 \hat{k}$, $\;$ $- \hat{i} + 2\hat{j} - \hat{k}$ act at a point $P$ whose position vector is $4 \hat{i} - 3 \hat{j} - 2 \hat{k}$. Find the moment of the resultant of the three forces acting at $P$ about the point $Q$ whose position vector is $6 \hat{i} + \hat{j} - 3 \hat{k}$.
Let $\overrightarrow{F_1} = 2 \hat{i} + 7 \hat{j}$; $\;$ $\overrightarrow{F_2} = 2 \hat{i} - 5 \hat{j} + 6 \hat{k}$; $\;$ $\overrightarrow{F_3} = - \hat{i} + 2 \hat{j} - \hat{k}$
$\begin{aligned}
\therefore \; \text{Resultant force} = \overrightarrow{F} & = \overrightarrow{F_1} + \overrightarrow{F_2} + \overrightarrow{F_3} \\\\
& = \left(2 \hat{i} + 7 \hat{j}\right) + \left(2 \hat{i} - 5 \hat{j} + 6 \hat{k}\right) + \left(- \hat{i} + 2 \hat{j} - \hat{k}\right) \\\\
& = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}
\end{aligned}$
Position vector of point $P = 4 \hat{i} - 3 \hat{j} - 2 \hat{k}$ $\;\;\;$ i.e. point $P \left(4, -3, -2\right)$
Position vector of point $Q = 6 \hat{i} + \hat{j} - 3 \hat{k}$ $\;\;\;$ i.e. point $Q \left(6, 1, -3\right)$
$\begin{aligned} \overrightarrow{QP} = \overrightarrow{r} & = \text{position vector of P} - \text{position vector of Q} \\\\ & = \left(4 \hat{i} - 3 \hat{j} - 2 \hat{k}\right) - \left(6 \hat{i} + \hat{j} - 3 \hat{k}\right) \\\\ & = - 2 \hat{i} - 4 \hat{j} + \hat{k} \end{aligned}$
Moment of force (or torque) acting at point $P$ about point $Q$ is
$\begin{aligned} \overrightarrow{M} = \overrightarrow{r} \times \overrightarrow{F} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & - 4 & 1 \\ 3 & 4 & 5 \end{vmatrix} \\\\ & = \hat{i} \left(- 20 - 4\right) - \hat{j} \left(- 10 - 3\right) + \hat{k} \left(- 8 + 12\right) \\\\ & = - 24 \hat{i} + 13 \hat{j} + 4 \hat{k} \end{aligned}$