Three point charges $+4e$ , $+2e$ and $+e$ are kept at each vertex of an equilateral triangle ABC , each of side $a$ . The net electric intensity acting on charge $+e$ is
( $\cos 60^\circ = 1/2$ and $k$ is constant )

Let the charge $+e$ be placed at vertex C of the equilateral triangle ABC, while $+4e$ and $+2e$ are placed at vertices A and B respectively.

The electric intensity (electric field) at C due to the charge $+4e$ at A is:
$E_1 = \dfrac{k(4e)}{a^2}$

The electric intensity at C due to the charge $+2e$ at B is:
$E_2 = \dfrac{k(2e)}{a^2}$

Since ABC is an equilateral triangle, the angle between the electric field vectors $\vec{E}_1$ and $\vec{E}_2$ is $60^\circ$.

The net electric intensity at C is given by the vector sum of $\vec{E}_1$ and $\vec{E}_2$:
$E_{net} = \sqrt{E_1^2 + E_2^2 + 2E_1 E_2 \cos 60^\circ}$

Substituting the values of $E_1$, $E_2$, and $\cos 60^\circ = \dfrac{1}{2}$:
$E_{net} = \sqrt{\left(\dfrac{4ke}{a^2}\right)^2 + \left(\dfrac{2ke}{a^2}\right)^2 + 2\left(\dfrac{4ke}{a^2}\right)\left(\dfrac{2ke}{a^2}\right)\left(\dfrac{1}{2}\right)}$

$E_{net} = \dfrac{ke}{a^2} \sqrt{16 + 4 + 8}$

$E_{net} = \dfrac{ke}{a^2} \sqrt{28}$

$E_{net} = \dfrac{2\sqrt{7}\,ke}{a^2}$

Answer: $\dfrac{2\sqrt{7}\,ke}{a^2}$