Three charges of equal magnitude $q$ are placed at the vertices of an equilateral triangle of side $l.$ The force on a charge $Q$ placed at the centroid of the triangle is

As shown in figure, draw $AD\perp BC.$

Therefore $AD=AB \cos {30}^{°}=\frac{l\sqrt{3}}{2}$

Distance (AO) of the centroid O from A

$=\frac{2}{3}AD=\frac{2l}{3}\frac{\sqrt{3}}{2}=\frac{l}{\sqrt{3}}$

Force on $Q$ at $O$ due to charge $q_1=q$ at $A,$

${\overrightarrow{F}}_1=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{Qq}{{(l/\sqrt{3})}^2}=\frac{3Qq}{4{\mathrm{\pi \epsilon }}_0{\mathrm{l}}^2},$along $AO$

Similarly, force on $O$ due to charge $q_2=q$ at $B,$

${\overrightarrow{F}}_2=\frac{3Qq}{4{\mathrm{\pi \epsilon }}_0{\mathrm{l}}^2},$along $BO$

and force on $Q$ due to charge $q_3=q$ at $C$

${\overrightarrow{F}}_3=\frac{3Qq}{4{\mathrm{\pi \epsilon }}_0{\mathrm{l}}^2},$along CO

Angle between forces $F_2$ and  $F_3={120}^{°}$

By parallelogram law, resultant of forces ${\overrightarrow{F}}_2$ and ${\overrightarrow{F}}_3$

$=\frac{3Qq}{4{\mathrm{\pi \epsilon }}_0{\mathrm{l}}^2},$along OA

Therefore Total force on $Q=\frac{3Qq}{4{\mathrm{\pi \epsilon }}_0{\mathrm{l}}^2}-\frac{3Qq}{4{\mathrm{\pi \epsilon }}_0{\mathrm{l}}^2}=0$

Alternate solution:

As the three charges are at equal distance from centroid and have equal magnitude of charge.

So the magnitude of forces is also equal, at they subtend equal angles.

So by lami's theorem we can say that the charge placed at centroid of the triangle will be at equilibrium.

So the net force is zero.