Answer
$$
\begin{aligned}
& \mathbf{M}_C=\{-2 \mathbf{i}+20 \mathbf{j}+17 \mathbf{k}\} \mathrm{kN} \cdot \mathrm{m} \\
& M_C=26.3 \mathrm{kN} \cdot \mathrm{m}
\end{aligned}
$$
Work Step by Step
The coordinates of points $A$ and $B$ are $A(0,0,1) \mathrm{m}$ and $B(3,2,-1) \mathrm{m}$, respectively. Thus,
$$
\mathbf{r}_{A B}=(3-0) \mathbf{i}+(2-0) \mathbf{j}+(-1-1) \mathbf{k}=\{3 \mathbf{i}+2 \mathbf{j}-2 \mathbf{k}\} \mathrm{m}
$$
Couple Moment.
$$
\begin{aligned}
\mathbf{M}_C & =\mathbf{r}_{A B} \times \mathbf{F} \\
& =\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
3 & 2 & -2 \\
-4 & 3 & -4
\end{array}\right| \\
& =\{-2 \mathbf{i}+20 \mathbf{j}+17 \mathbf{k}\} \mathrm{kN} \cdot \mathrm{m}
\end{aligned}
$$
The magnitude of $\mathbf{M}_C$ is
$$
\begin{aligned}
M_C & =\sqrt{\left(M_C\right)_x^2+\left(M_C\right)_y^2+\left(M_C\right)_z^2} \\
& =\sqrt{(-2)^2+20^2+17^2} \\
& =26.32 \mathrm{kN} \cdot \mathrm{m}=26.3 \mathrm{kN} \cdot \mathrm{m}
\end{aligned}
$$
