Answer
$$
\mathbf{M}_A=\{574 \mathbf{i}+350 \mathbf{j}+1385 \mathbf{k}\} \mathrm{N} \cdot \mathbf{m}
$$
Work Step by Step
The coordinates of points $A, B$ and $C$ are $A(0,0,4) \mathrm{m}, B\left(4 \sin 45^{\circ}, 0,4 \cos 45^{\circ}\right) \mathrm{m}$ and $C(6,6,0) \mathrm{m}$, respectively. Thus
$$
\begin{aligned}
\mathbf{r}_{A B} & =\left(4 \sin 45^{\circ}-0\right) \mathbf{i}+(0-0) \mathbf{j}+\left(4 \cos 45^{\circ}-4\right) \mathbf{k} \\
& =\{2.8284 \mathbf{i}-1.1716 \mathbf{k}\} \mathbf{m} \\
\mathbf{r}_{A C} & =(6-0) \mathbf{i}+(6-0) \mathbf{j}+(0-4) \mathbf{k} \\
& =\{6 \mathbf{i}+6 \mathbf{j}-4 \mathbf{k}\} \mathbf{m} \\
\mathbf{r}_{B C} & =\left(6-4 \sin 45^{\circ}\right) \mathbf{i}+(6-0) \mathbf{j}+\left(0-4 \cos 45^{\circ}\right) \mathbf{k} \\
& =\{3.1716 \mathbf{i}+6 \mathbf{j}-2.8284 \mathbf{k}\} \mathbf{m} \\
\mathbf{F} & =F\left(\frac{\mathbf{r}_{B C}}{\mathbf{r}_{B C}}\right)=600\left(\frac{3.1716 \mathbf{i}+6 \mathbf{j}-2.8284 \mathbf{k}}{\sqrt{3.1716^2+6^2+(-2.8284)^2}}\right) \\
& =\{2.58 .82 \mathbf{i}+489.63 \mathbf{j}-230.81 \mathbf{k}\} \mathrm{N}
\end{aligned}
$$
The Moment of Force $F$ About Point $A$.
$$
\begin{aligned}
\mathbf{M}_A & =\mathbf{r}_{A B} \times \mathbf{F} \\
& =\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2.8284 & 0 & -1.1716 \\
258.82 & 489.63 & -230.81
\end{array}\right| \\
& =\{573.64 \mathbf{i}+349.62 \mathbf{j}+1384.89 \mathbf{k}\} \mathrm{N} \cdot \mathrm{m} \\
& =\{574 \mathbf{i}+350 \mathbf{j}+1385 \mathbf{k}\} \mathbf{N} \cdot \mathrm{m}
\end{aligned}
$$
