hi,

potential due to three chrges at a point is= -kq/r1-kq/r2-kq/r3.

thus potential=-kq(1/r1+1/r2+1/r3)

if potential has to be zero then that point should be at infinity ( r1r2r3 should be at infinity)

plzz approve if you like the answer

dear shiven , first u draw the figure according to these steps

1)draw a equilateral triangle ABC such that its base AB is on x axis , center of base is at origin...
third vertice (C) is on y axis ...

2)y axis divides the base in two halves , AO ,BO ...

3) consider any point P(0,y) on y axis then

4) place -q , -q at A,B & q at C...

now , potential at P will be

V = k(-q)/AP + K-(q)/BP + kq/PC .......1

AO = BO = a/2

PC = asin60 - y = [aroot3/2 - y] ........2

AP = BP = [y^2 + a^2/4]^1/2 ......3 ( by pythagorus theorem)

for potential to become 0 , eq 1 = 0 so

q/AP + q/BP = q/PC

2/AP = 1/PC

AP = 2pC .........4

putting values of AP,PC from eq 2 & 3 we get a quadratic expression

3y^2 - (4root3a)y + 11a^2/4 = 0

from here Y = a[4root3 +(-) root15]/6

= a[ 4root3 - root15]/6 or [4root3 + root15]/6 ( not acceptable coz this point lies ouside)

therefore the only point is Y = a[4root3-root15]/6

this is the required point where potential will be zero ...