Moment of a force of magnitude $20 \, N$ acting along the positive $x$-direction at point $(3 \, m, 0, 0)$ about the point $(0, 2, 0)$ (in $N \cdot m$) is ...........

The vector sum of a system of non-collinear forces acting on a rigid body is given to be nonzero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero,does this mean that it is necessarily zero about any arbitrary point?

Which type of motion exists due to the moment of force (couple)?

$A$ particle of mass $m$ is moving along a trajectory given by $x = x_0 + a \cos \omega_1 t$ and $y = y_0 + b \sin \omega_2 t$. The torque acting on the particle about the origin at $t = 0$ is:

The center of an equilateral triangle $ABC$ is $O$. Three forces $F_1$,$F_2$,and $F_3$ are applied along the sides $AB$,$BC$,and $AC$ respectively. If the total torque about $O$ is zero,what is the relationship between $F_1$,$F_2$,and $F_3$?

How is the direction of torque determined?