Answer
$$
\begin{aligned}
& M_a=4.37 \mathrm{~N} \cdot \mathrm{m} \\
& \alpha=33.7^{\circ} \\
& \beta=90^{\circ} \\
& \gamma=56.3^{\circ} \\
& M=5.41 \mathrm{~N} \cdot \mathrm{m}
\end{aligned}
$$
Work Step by Step
$$
\begin{aligned}
\mathbf{F} & =30\left(\cos 60^{\circ} \mathbf{i}+\cos 60^{\circ} \mathbf{j}+\cos 45^{\circ} \mathbf{k}\right) \\
& =\{15 \mathbf{i}+15 \mathbf{j}+21.21 \mathbf{k}\} \mathrm{N} \\
\mathbf{r} & =\{-0.1 \mathbf{i}+0.15 \mathbf{k}\} \mathrm{m} \\
\mathbf{u} & =\mathbf{j} \\
M_\alpha & =\left|\begin{array}{ccc}
0 & 1 & 0 \\
-0.1 & 0 & 0.15 \\
15 & 15 & 21.21
\end{array}\right|=4.37 \mathrm{~N} \cdot \mathrm{m}
\end{aligned}
$$
F must be perpendicular to $\mathbf{u}$ and $\mathbf{r}$.
$$
\begin{aligned}
\mathbf{u}_f & =\frac{0.15}{0.1803} \mathbf{i}+\frac{0.1}{0.1803} \mathbf{k} \\
& =0.8321 \mathbf{i}+0.5547 \mathbf{k} \\
\alpha & =\cos ^{-1} 0.8321=33.7^{\circ} \\
\beta & =\cos ^{-1} 0=90^{\circ} \\
\gamma & =\cos ^{-1} 0.5547=56.3^{\circ} \\
M & =30(0.1803)\\&=5.41 \mathrm{~N} \cdot \mathrm{m}
\end{aligned}
$$
