The moment of the force, $\overrightarrow{\mathrm{F}}=4\hat{\mathrm{i}}+5\hat{\mathrm{j}}-6\hat{\mathrm{k}}$ at (2, 0, -3), about the point (2, -2, -2), is given by:

$\text{ Moment of the force is, }\overrightarrow{\tau }=\left(\overrightarrow{r}-{\overrightarrow{r}}_0\right)\times \overrightarrow{F}$

$\text{ Here, }{\overrightarrow{r}}_0=2\hat{i}-2\hat{j}-2\hat{k}\text{ and }\overrightarrow{r}=2\hat{i}+0\hat{j}-3\hat{k}$

$\therefore \overrightarrow{r}-{\overrightarrow{r}}_0=(2\hat{i}+0\hat{j}-3\hat{k})-(2\hat{i}-2\hat{j}-2\hat{k})=0\hat{i}+2\hat{j}-\hat{k}$

$\therefore \overrightarrow{\tau }=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 2& -1\\ 4& 5& -6\end{array}\right|=-7\hat{i}-4\hat{j}-8\hat{k}$