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(C) Let the perpendicular distance from the center $O$ to each side of the equilateral triangle be $r$.The torque $	au$ due to a force $F$ is given b

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$F_{1}+F_{2}$

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(C) Let the perpendicular distance from the center $O$ to each side of the equilateral triangle be $r$.The torque $	au$ due to a force $F$ is given by $	au = F \cdot r$,where $r$ is the perpendicular distance from the axis of rotation to the line of action of the force.Looking at the directions of the forces in the figure,the forces $\overrightarrow{F}_{1}$ and $\overrightarrow{F}_{2}$ create a torque in the same rotational sense (e.g.,clockwise) about point $O$,while $\overrightarrow{F}_{3}$ creates a torque in the opposite sense (e.g.,counter-clockwise).For the total torque about point $O$ to be zero,the sum of the torques must be zero:$	au_{1} + 	au_{2} - 	au_{3} = 0$$rF_{1} + rF_{2} - rF_{3} = 0$Dividing by $r$ (since $r 
eq 0$):$F_{1} + F_{2} - F_{3} = 0$$F_{3} = F_{1} + F_{2}$

The figure shows three forces $\overrightarrow{F}_{1}, \overrightarrow{F}_{2}$ and $\overrightarrow{F}_{3}$ acting along the sides of an equilateral triangle. If the total torque acting at point $O$ (the center of the triangle) is zero,then the magnitude of $\overrightarrow{F}_{3}$ is:

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