Let $r$ be the perpendicular distance from the centre O of the equilateral triangle to each of its sides. The direction of torque is determined by the cross product of the position vector and the force vector. Forces $\vec{F}_1$ (acting along AB) and $\vec{F}_2$ (acting along BC) will produce torque in the same rotational sense (either both clockwise or both counter-clockwise) about the centre O. However, the force $\vec{F}_3$ is acting along AC (not CA, which would be the continuous cyclic order). Therefore, $\vec{F}_3$ will produce a torque in the opposite sense to that of $\vec{F}_1$ and $\vec{F}_2$. The total torque about O is the algebraic sum of the individual torques: $\tau_{\text{total}} = F_1 r + F_2 r - F_3 r$ Given that the total torque about O is zero: $F_1 r + F_2 r - F_3 r = 0$ Dividing by $r$ (since $r \neq 0$): $F_3 = F_1 + F_2$