Answer
$y=1$
Work Step by Step
Form an equation using the distance formula
(distance from (0,0,0) to (x,y,z))=(distance from (0,2,0) to (x,y,z))
$\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{x^{2}+(y-2)^{2}+z^{2}}\quad $ ... square both sides
$x^{2}+y^{2}+z^{2}=x^{2}+(y-2)^{2}+z^{2}\quad $ ... simplify
$y^{2}=y^{2}-4y+4$
$4y=4$
$y=1$
(the plane parallel to the xz plane, containing (0,1,0)).
