Answer
$$
\mathbf{M}_A=\{-5.39 \mathbf{i}+13.1 \mathbf{j}+11.4 \mathbf{k}\} \mathbf{N} \cdot \mathrm{m}
$$
Work Step by Step
Position Vector And Force Vector:
$$
\begin{aligned}
\mathbf{r}_{A C} & =\{(0.55-0) \mathbf{i}+(0.4-0) \mathbf{j}+(-0.2-0) \mathbf{k}\} \mathbf{m} \\
& =\{0.55 \mathbf{i}+0.4 \mathbf{j}-0.2 \mathbf{k}\} \mathbf{m} \\
\mathbf{F} & =80\left(\cos 30^{\circ} \sin 40^{\circ} \mathbf{i}+\cos 30^{\circ} \cos 40^{\circ} \mathbf{j}-\sin 30^{\circ} \mathbf{k}\right) \mathrm{N} \\
& =(44.53 \mathbf{i}+53.07 \mathbf{j}-40.0 \mathbf{k}\} \mathrm{N}
\end{aligned}
$$
Moment of Force F About Point A: Applying Eq. 4-7, we have
$$
\begin{aligned}
\mathbf{M}_A & =\mathbf{r}_{A C} \times \mathbf{F} \\
& =\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0.55 & 0.4 & -0.2 \\
44.53 & 53.07 & -40.0
\end{array}\right| \\
& =|-5.39 \mathbf{i}+13.1 \mathbf{j}+11.4 \mathbf{k}| \mathbf{N} \cdot \mathbf{m}
\end{aligned}
$$
