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)
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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 ...