Answer
$$
\mathbf{M}_A=\{-82.9 \mathbf{i}+41.5 \mathbf{j}+232 \mathbf{k}\} \mathbf{l b} \cdot \mathrm{ft}
$$
Work Step by Step
The coordinates of points $A, C$ and $D$ are $A(-6.5,-3,0) \mathrm{ft}, C\left[0,-\left(3+4 \cos 45^{\circ}\right), 4 \sin 45^{\circ}\right]$ ft and $D(-5,0,0) \mathrm{ft}$, respectively. Thus,
$$
\begin{aligned}
\mathbf{r}_{A C} & =[0-(-6.5)] \mathbf{i}+\left[-\left(3+4 \cos 45^{\circ}\right)-(-3)\right] \mathbf{j}+\left(4 \sin 45^{\circ}-0\right) \mathbf{k} \\
& =\{6.5 \mathbf{i}-2.8284 \mathbf{j}+2.8284 \mathbf{k}\} \mathbf{f t} \\
\mathbf{r}_{A D} & =[-5-(-6.5)] \mathbf{i}+[0-(-3)] \mathbf{j}+(0-0) \mathbf{k}=\{1.5 \mathbf{i}+3 \mathbf{j}\} \mathrm{ft} \\
\mathbf{r}_{C D} & =(-5-0) \mathbf{i}+\left\{0-\left[-\left(3+4 \cos 45^{\circ}\right)\right]\right\} \mathbf{j}+\left(0-4 \sin 45^{\circ}\right) \mathbf{k} \\
& =\{-5 \mathbf{i}+5.8284 \mathbf{j}-2.8284 \mathbf{k}\} \mathrm{ft} \\
\mathbf{F} & =F\left(\frac{\mathbf{r}_{C D}}{\mathbf{r}_{C D}}\right)=80\left(\frac{-5 \mathbf{i}+5.8284 \mathbf{j}-2.8284 \mathbf{k}}{\sqrt{(-5)^2+5.8284^2+(-2.8284)^2}}\right) \\
& =\{-48.88 \mathbf{i}+56.98 \mathbf{j}-27.65 \mathbf{k}\} \mathbf{l b}
\end{aligned}
$$
Moment of $F$ About Point $A$.
$$
\begin{aligned}
\mathbf{M}_A & =\mathbf{r}_{A C} \times \mathbf{F} \\
& =\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
6.5 & -2.8284 & 2.8284 \\
-48.88 & 56.98 & -27.65
\end{array}\right| \\
& =\{-82.9496 \mathbf{i}+41.47 \mathbf{j}+232.10 \mathbf{k}\} \mathbf{l b} \cdot \mathrm{ft} \\
& =\{-82.9 \mathbf{i}+41.5 \mathbf{j}+232 \mathbf{k}\} \mid \mathrm{b} \cdot \mathrm{ft}
\end{aligned}
$$
