Let $A=(1,0),B(-1,0),C(2,0)$ then the locus of point P such that $P{B}^2+P{C}^2=2P{A}^2$ is

Given $A=(1,0),B(-1,0),C(2,0)$, $P{B}^2+P{C}^2=2P{A}^2$

Locus of point $P(x,y)$ such that

${(x+1)}^2+y^2+{(x-2)}^2+y^2=2[{(x-1)}^2+y^2]$

$x^2+2x+1+y^2+x^2-4x+4+y^2=2[x^2-2x+1+y^2]$

$2x=-3$

$x=\frac{-3}{2}$

The straight line which is parallel to y-axis.