If force $$\vec{F} = 3\hat{i} + 4\hat{j} - 2\hat{k}$$ acts on a particle having position vector $$2\hat{i} + \hat{j} + 2\hat{k}$$ then, the torque about the origin will be:

We need to find the torque about the origin given $$\vec{F} = 3\hat{i} + 4\hat{j} - 2\hat{k}$$ and position vector $$\vec{r} = 2\hat{i} + \hat{j} + 2\hat{k}$$.

The torque is given by the cross product of $$\vec{r}$$ and $$\vec{F}$$: $$\vec{\tau} = \vec{r} \times \vec{F}$$.

Using the determinant method, we compute $$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 3 & 4 & -2 \end{vmatrix}$$.

Expanding this determinant gives the components: the $$\hat{i}$$ component is $$(1)(-2) - (2)(4) = -2 - 8 = -10$$; the $$\hat{j}$$ component is $$-[(2)(-2) - (2)(3)] = -[-4 - 6] = 10$$; and the $$\hat{k}$$ component is $$(2)(4) - (1)(3) = 8 - 3 = 5$$.

Therefore, $$\vec{\tau} = -10\hat{i} + 10\hat{j} + 5\hat{k}$$.

The correct answer is Option A: $$-10\hat{i} + 10\hat{j} + 5\hat{k}$$.