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In the given equilateral triangle ABC of sides of length l, if we draw a perpendicular AD to the side $BC,AD = AC \cos 30 ^{\circ} = ( \sqrt 3 /2) l$ and the distance AO of the centroid O from A is $(2/3)AD =(1/ \sqrt 3) l$.

By symmetry $AO=BO=CO$

Thus,

Force $F_1$ on Q due to charge q at $A =\large\frac{3}{4 \pi \in_0} \frac{Qq}{l^2}$ along $AO$

Force $F_2$ on Q due to charge q at $B =\large\frac{3}{4 \pi \in_0} \frac{Qq}{l^2}$ along $BO$

Force $F_3$ on Q due to charge q at $C =\large\frac{3}{4 \pi \in_0} \frac{Qq}{l^2}$ along $CO$

The resultant of forces $F_2 $ and $F_3$ is $\large\frac{3}{4 \pi \in_0} \frac{Qq}{l^2} $$( \hat {r} - \hat {r})=0$ where $\hat {r}4$ is the unit vector along OA.

It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but in some direction.

Consider what would happen if the system was rotated through $60^{\circ}$ about O.

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