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

<p>First, torque ($\vec{\tau}$) is found by taking the cross product of the position vector ($\vec{r}$) and the force vector ($\vec{F}$):</p> <p> $\vec{\tau} = \vec{r} \times \vec{F}$ </p> <p> Here, $\vec{r} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{F} = 2\hat{i} + \hat{j} + 2\hat{k}$. </p> <p> We use a determinant to find the cross product: $$ \vec{\tau} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 1 & 2 \\ \end{array} \right| $$ </p> <p> Expanding the determinant: </p> <p> $\vec{\tau} = \hat{i}[(1)(2)-(1)(1)] -\hat{j}[(1)(2)-(1)(2)] + \hat{k}[(1)(1)-(1)(2)]$ </p> <p> This gives: $\vec{\tau} = \hat{i}(2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 - 2)$ </p> <p> So, $\vec{\tau} = \hat{i} - 0\hat{j} - \hat{k}$ </p>