Find the coordinates of a point on $y$-axis which are at a distance of $5\sqrt{2}$ from the point $P(3,-2,5)$

Let $A(0,b,0)$ be the point on the $y$-axis at a distance of $5\sqrt{2}$ from point  $P(3,-2,5)$.
We have given, $AP=5\sqrt{2}$
$\Rightarrow A{P}^2=50$
$\Rightarrow {\left(3-0\right)}^2+{\left(-2-B\right)}^2+{\left(5-0\right)}^2=50$
$\Rightarrow 9+4+b^2+4b+25=50$
$\Rightarrow b^2+4b-12=0$
$\Rightarrow b^2+6b-2b-12=0$
$\Rightarrow \left(b+6\right)\left(b-2\right)=0$
$\Rightarrow b=-6,2$
Thus, the coordinates of the required points are $(0,2,0)$ and $(0,-6,0)$