There are three forces $\vec{F_1}$,$\vec{F_2}$,and $\vec{F_3}$ acting on a body,all acting on a point $P$ on the body. The body is found to move with uniform speed.
$(a)$ Show that the forces are coplanar.
$(b)$ Show that the torque acting on the body about any point due to these three forces is zero.

(N/A) Since the body moves with uniform speed (constant velocity),its acceleration is zero,i.e.,$\vec{a} = 0$. According to Newton's second law,the net force acting on the body is $\vec{F}_{net} = m\vec{a} = 0$. Thus,$\vec{F_1} + \vec{F_2} + \vec{F_3} = 0$.
$(a)$ Since the sum of the three forces is zero,they must form a closed triangle when placed head-to-tail. $A$ triangle is a two-dimensional figure,meaning all three vectors must lie in the same plane. Therefore,the forces are coplanar.
$(b)$ The torque $\vec{\tau}$ about any point $O$ is given by $\vec{\tau} = \sum (\vec{r_i} \times \vec{F_i})$. Since all forces act at point $P$,the position vector of the point of application for all forces relative to $P$ is zero. Thus,the torque about point $P$ is zero. For any other point $O$,the total torque is $\vec{\tau}_O = \vec{OP} \times (\vec{F_1} + \vec{F_2} + \vec{F_3})$. Since $\vec{F_1} + \vec{F_2} + \vec{F_3} = 0$,the torque about any point $O$ is also zero.