Question:
Find the coordinates of a point on y-axis which are at a distance of $5 \sqrt{2}$ from the point $P(3,-2,5)$.
Solution:
If a point is on the y-axis, then x-coordinate and the z-coordinate of the point are zero.
Let $\mathrm{A}(0, b, 0)$ be the point on the $y$-axis at a distance of $5 \sqrt{2}$ from point $\mathrm{P}(3,-2,5)$. Accordingly, $\mathrm{AP}=5 \sqrt{2}$
$\therefore \mathrm{AP}^{2}=50$
$\Rightarrow(3-0)^{2}+(-2-b)^{2}+(5-0)^{2}=50$
$\Rightarrow 9+4+b^{2}+4 b+25=50$
$\Rightarrow b^{2}+4 b-12=0$
$\Rightarrow b^{2}+6 b-2 b-12=0$
$\Rightarrow(b+6)(b-2)=0$
$\Rightarrow b=-6$ or 2
Thus, the coordinates of the required points are (0, 2, 0) and (0, –6, 0).