Three charges are placed at the vertices of an equilateral triangle...

The force experienced by $A$ due to $B$ is $F_1=\frac{1}{4 \pi \varepsilon_0} \frac{Q Q}{a^2}$ along $\overrightarrow{A B}$ (attractive) The force experienced by $A$ due to $C$ is $F_2=\frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{a^2}$ along $\overrightarrow{C A}$ produced (repulsive)
Now the force $\vec{F}_1$ is having a component $F_1 \cos 30^{\circ}$ along and the force $\vec{F}_2$ is having a component $F_2 \cos 30^{\circ}$ along $\overrightarrow{A E}, \overrightarrow{A E} and$\overrightarrow{A F}$are both normal to$\overrightarrow{B C}$but they are mutually opposite to each other. And as \left|\vec{F}_1\right|=\left|\vec{F}_2\right|$, so the forces along $\overrightarrow{A E}$ and $\overrightarrow{A F}$ will both cancel each other. And so the force experienced by the charge at $A$ in the direction normal to $\overrightarrow{B C}$ is zero. $$ \begin{aligned}