To find the moment of the force $\vec{F}$ about a point, we use the formula for the moment (torque):$\\ \vec{\tau} = \vec{r} \times \vec{F} \\$where $\vec{r}$ is the position vector from the point about which the moment is calculated to the point of application of the force.$\\$Given:$\\ \vec{F} = 4 \hat{i} + 5 \hat{j} - 6 \hat{k} \\$Point of application of the force: $(2, 0, -3) \\$Point about which moment is calculated: $(2, -2, -2) \\$First, calculate the position vector $\vec{r}: \\ \vec{r} = (2 - 2) \hat{i} + (0 + 2) \hat{j} + (-3 + 2) \hat{k} \\ \vec{r} = 0 \hat{i} + 2 \hat{j} - 1 \hat{k} \\$Now, calculate the cross product $\vec{\tau} = \vec{r} \times \vec{F}: \\ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & -1 \\ 4 & 5 & -6 \\ \end{vmatrix} \\$Calculate the determinant:$\\ \vec{\tau} = \hat{i} \left( (2)(-6) - (-1)(5) \right) - \hat{j} \left( (0)(-6) - (-1)(4) \right) + \hat{k} \left( (0)(5) - (2)(4) \right) \\ \vec{\tau} = \hat{i} (-12 + 5) - \hat{j} (0 + 4) + \hat{k} (0 - 8) \\ \vec{\tau} = -7 \hat{i} - 4 \hat{j} - 8 \hat{k} \\$Therefore, the moment of the force about the point is $-7 \hat{i} - 4 \hat{j} - 8 \hat{k}. \\$This corresponds to Option 4.