In the given equilateral triangle $ABC$ of sides of length $l$, if Iraw a perpendicular $AD$ to the side $BC,$

$A D=A C \cos 30^{\circ}=(\sqrt{3} / 2) l$ and the distance $AO$ of the centroid $O$ from $A$ is $(2 / 3) AD =(1 / \sqrt{3})$ $l$. By symmatry $AO = BO = CO$

Thus,

Force $F _{1}$ on $Q$ due to charge $q$ at $A =\frac{3}{4 \pi \varepsilon_{0}} \frac{ Q q}{l^{2}}$ along $AO$

Force $F _{2}$ on $Q$ due to charge $q$ at $B =\frac{3}{4 \pi \varepsilon_{0}} \frac{ Q q}{l^{2}}$ along $BO$

Force $F_{3}$ on $Q$ due to charge $q$ at $C=\frac{3}{4 \pi \varepsilon_{0}} \frac{Q q}{l^{2}}$ along $CO$

The resultant of forces $F _{2}$ and $F _{3}$ is $\frac{3}{4 \pi \varepsilon_{0}} \frac{Q q}{l^{2}}$ along $OA$. by the parallelogram law. Therefore, the total force on $g=\frac{3}{4 \pi \varepsilon_{0}} \frac{Q q}{l^{2}}(\hat{ r }-\hat{ r })$

$=0,$ where $\hat{ r }$ 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 $O$.