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

Moment of inertia

$\stackrel{\rightharpoonup }{\mathrm{\tau }} = \overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{F}}$

$\overrightarrow{\mathrm{r}} = (2-2)\hat{\mathrm{i}}+(0-(-2\left)\right)\hat{\mathrm{j}}+\left(\right(-3)-(-2\left)\right)\hat{\mathrm{k}}$

$= (2\hat{\mathrm{j}}-\hat{\mathrm{k}})$

Hence torque of force F about O

$\overrightarrow{\mathrm{\tau }} = (2\hat{\mathrm{j}}-\hat{\mathrm{k}})\times (4\hat{\mathrm{i}}+5\hat{\mathrm{j}}-6\hat{\mathrm{k}}) = -7\hat{\mathrm{i}}-4\hat{\mathrm{j}}-8\hat{\mathrm{k}}$