Let the point of application of the force be $P$ and the reference point be $O$. The position vector of the point of application is $\vec{r}_P = 2\hat{i} + 0\hat{j} - 3\hat{k}$. The position vector of the point about which the torque is to be calculated is $\vec{r}_O = 2\hat{i} - 2\hat{j} - 2\hat{k}$.

The position vector $\vec{r}$ of the point of application with respect to the reference point is: $\vec{r} = \vec{r}_P - \vec{r}_O = (2 - 2)\hat{i} + (0 - (-2))\hat{j} + (-3 - (-2))\hat{k} = 0\hat{i} + 2\hat{j} - \hat{k}$.

The torque (moment of force) $\vec{\tau}$ is given by the cross product of $\vec{r}$ and $\vec{F}$: $\vec{\tau} = \vec{r} \times \vec{F}$ $\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\0 & 2 & -1 \\4 & 5 & -6 \end{vmatrix}$ $\vec{\tau} = \hat{i}[(2)(-6) - (-1)(5)] - \hat{j}[(0)(-6) - (-1)(4)] + \hat{k}[(0)(5) - (2)(4)]$ $\vec{\tau} = \hat{i}(-12 + 5) - \hat{j}(0 + 4) + \hat{k}(0 - 8)$ $\vec{\tau} = -7\hat{i} - 4\hat{j} - 8\hat{k}$.