The torque due to the force $$(2\widehat{i}+\widehat{j}+2\widehat{k})$$ about the origin, acting on a particle whose position vector is $$(\widehat{i}+\widehat{j}+\widehat{k})$$, would be

The torque $$\vec{\tau}$$ is given by $$\vec{\tau} = \vec{r} \times \vec{F}$$, where $$\vec{r} = \hat{i} + \hat{j} + \hat{k}$$ and $$\vec{F} = 2\hat{i} + \hat{j} + 2\hat{k}$$.

Computing the cross product using the determinant formula:

$$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 1 & 2 \end{vmatrix}$$

$$\hat{i}$$ component: $$(1)(2) - (1)(1) = 2 - 1 = 1$$

$$\hat{j}$$ component: $$-[(1)(2) - (1)(2)] = -(2 - 2) = 0$$

$$\hat{k}$$ component: $$(1)(1) - (1)(2) = 1 - 2 = -1$$

$$\vec{\tau} = \hat{i} - \hat{k}$$

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