Answer
The proof is below.
Work Step by Step
We find the sum of the torques about point P:
$\tau_p=(F_{1p}\times r_{1p})+(F_{2p}\times r_{2p})+(F_{3p}\times r_{3p})$
Using substitution, we find:
$\tau_p=\vec{0}+\vec{0}\times R = \vec{0}$
We now consider point O:
$\tau_O=(F_{1O}\times r_{1O})+(F_{2O}\times r_{2O})+(F_{3O}\times r_{3O})$
$\tau_O=\vec{0}$
Thus, regardless of the pivot point, the net torque is 0.
