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Force due to two other point charges on the third point charge will be along the sides joining them.  But away from them outwards along the sides of the triangle.

the forces make angle 30 deg. with the altitude of the triangle.

F_1 = F_2= \frac{1}{4 \pi \epsilon_0 }\frac{q^2}{a^2}\\\\vector\ sum\ of\ F_1\ and\ F_2:\\\\=F_1\ Cos\ 30^0+F_2\ Cos\ 30^0\\\\=\frac{\sqrt{3}}{4 \pi \epsilon_0 }\frac{q^2}{a^2}

the direction of th  result ant force is along the altitude at that vertex, away from the triangle.

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click on thanks butto n above pls