Answer
$(2,1,\sqrt {5}), (2,1,-\sqrt {5})$
Work Step by Step
From the question, we have $z=\sqrt {x^2+y^2}$ ....(1)
Distance $d=\sqrt{x-l)^2+(y-m)^2+(z-n)^2}$
Thus, $d=\sqrt{(x-4)^2+(y-2)^2+(x^2+y^2)}$ [From equation (1)]
$F=d^2=(x-4)^2+(y-2)^2+(x^2+y^2)$
This implies, $F_x=4x-8, F_y=4y-4$
After solving for $x$ and $y$, we get
$x=2$ and $y=1$
Thus, $z=x^2+y^2=2^2+1^2=\pm \sqrt {5}$
The required points are: $(2,1,\sqrt {5}), (2,1,-\sqrt {5})$
