The moment of the force,$\overrightarrow{F} = 4\hat{i} + 5\hat{j} - 6\hat{k}$ at point $(2, 0, -3)$,about the point $(2, -2, -2)$,is given by

$A$ wheel of radius $R$ with an axle of radius $R/2$ is as shown in the figure. It is free to rotate about a frictionless axis through its centre and perpendicular to the page. Three forces are exerted as shown in the figure: a force $F$ at an angle of $45^{\circ}$ to the tangent of the outer rim,a force $F$ tangential to the inner axle,and a force $2F$ tangential to the outer rim. The magnitude of the net torque acting on the system is nearly ............. $FR$.

$A$ wheel of radius $R$ with an axle of radius $R/2$ is shown in the figure and is free to rotate about a frictionless axis through its centre and perpendicular to the page. Three forces are exerted as shown in the figure. The magnitude of the net torque acting on the system is nearly (in $FR$)

What is the torque of force $\vec F = 2\hat i - 3\hat j + 4\hat k$ acting at a point $\vec r = 3\hat i + 2\hat j + 3\hat k$ about the origin?

How is the direction of torque determined?

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