Answer
$$
\mathbf{M}_C=\{-35.4 \mathbf{i}-128 \mathbf{j}-222 \mathbf{k}\} \mathrm{lb} \cdot \mathrm{ft}
$$
Work Step by Step
Position Vector and Force Vector:
$$
\begin{aligned}
\mathbf{r}_{C A} & =\left\{\left(5 \sin 60^{\circ}-0\right) \mathbf{j}+\left(5 \cos 60^{\circ}-5\right) \mathbf{k}\right\} \mathrm{m} \\
& =\{4.330 \mathbf{j}-2.50 \mathbf{k}\} \mathrm{m} \\
\mathbf{F}_{A B} & =60\left(\frac{(6-0) \mathbf{i}+\left(7-5 \sin 60^{\circ}\right) \mathbf{j}+\left(0-5 \cos 60^{\circ}\right) \mathbf{k}}{\sqrt{(6-0)^2+\left(7-5 \sin 60^{\circ}\right)^2+\left(0-5 \cos 60^{\circ}\right)^2}}\right) 1 \mathrm{~b} \\
& =\{51.231 \mathbf{i}+22.797 \mathbf{j}-21.346 \mathbf{k}\} \mathrm{Ib}
\end{aligned}
$$
Moment of Force $\mathbf{F}_{A B}$ About Point $C$ : Applying Eq. 4-7, we have
$$
\begin{aligned}
\mathbf{M}_C & =\mathbf{r}_{C A} \times \mathbf{F}_{A B} \\
& =\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & 4.330 & -2.50 \\
51.231 & 22.797 & -21.346
\end{array}\right| \\
& =\{-35.4 \mathbf{i}-128 \mathbf{j}-222 \mathbf{k}\} \mathrm{lb} \cdot \mathbf{f t}
\end{aligned}
$$
