Answer
$$
M_x=440 \mathrm{lb} \cdot \mathrm{ft}
$$
Work Step by Step
Using $x^{\prime}, y^{\prime}, z$ :
$$
\begin{aligned}
& \mathbf{u}_x=\cos 30^{\circ} \mathbf{i}^{\prime}+\sin 30^{\circ} \mathbf{j}^{\prime} \\
& \mathbf{r}_{A C}=-6 \cos 15^{\circ} \mathbf{i}^{\prime}+3 \mathbf{j}^{\prime}+6 \sin 15^{\circ} \mathbf{k} \\
& \mathrm{F}=80 \mathbf{k} \\
& M_x=\left|\begin{array}{ccc}
\cos 30^{\circ} & \sin 30^{\circ} & 0 \\
-6 \cos 15^{\circ} & 3 & 6 \sin 15^{\circ} \\
0 & 0 & 80
\end{array}\right|=207.85+231.82+0 \\
& M_x=440 \mathrm{lb} \cdot \mathrm{ft}
\end{aligned}
$$
Now, using $x, y, z$,
The coordinates of point $C$ :
$$
\begin{aligned}
& x=3 \sin 30^{\circ}-6 \cos 15^{\circ} \cos 30^{\circ}=-3.52 \mathrm{ft} \\
& y=3 \cos 30^{\circ}+6 \cos 15^{\circ} \sin 30^{\circ}=5.50 \mathrm{ft} \\
& z=6 \sin 15^{\circ}=1.55 \mathrm{ft} \\
& \mathbf{r}_{A C}=-3.52 \mathbf{i}+5.50 \mathbf{j}+1.55 \mathbf{k} \\
& \mathbf{F}=80 \mathbf{k} \\
& M_x=\left|\begin{array}{ccc}
1 & 0 & 0 \\
-3.52 & 5.50 & 1.55 \\
0 & 0 & 80
\end{array}\right|=440 \mathrm{lb} \cdot \mathrm{ft}
\end{aligned}
$$
