Which physical quantity plays the same role in rotational motion as force does in linear motion?

The figure shows three forces $\overrightarrow{F}_{1}, \overrightarrow{F}_{2}$ and $\overrightarrow{F}_{3}$ acting along the sides of an equilateral triangle. If the total torque acting at point $O$ (the center of the triangle) is zero,then the magnitude of $\overrightarrow{F}_{3}$ is:

When a force of $6.0 \, N$ is exerted at $30^{\circ}$ to a wrench at a distance of $8 \, cm$ from the nut,it is just able to loosen the nut. What force $F$ would be sufficient to loosen it,if it acts perpendicularly to the wrench at $16 \, cm$ from the nut (in $, N$)?

If force $\vec{F} = -3 \hat{i} + \hat{j} + 5 \hat{k}$ acts at a position vector $\vec{r} = 7 \hat{i} + 3 \hat{j} + \hat{k}$,then the torque $\vec{\tau}$ acting at that point is:

The torque of a force $\overrightarrow{F} = -3\hat{i} + \hat{j} + 5\hat{k}$ acting at the point $\overrightarrow{r} = 7\hat{i} + 3\hat{j} + \hat{k}$ is:

$A$ force $F\,\hat{k}$ acts at the origin $O$ of a coordinate system. What is the torque about the point $(1, -1, 0)$?