Using vectors:
Find the point on the y axis that is equidistant from the points (2,-1,1) and (0,1,3)
Hi
The point $\displaystyle P$ we're looking for is on the $\displaystyle y$ axis hence $\displaystyle P(0,\,y,\,0)$
The distance between $\displaystyle P(0,\,y,\,0)$ and $\displaystyle A(2,\,-1,\,1)$ is $\displaystyle \|\vec{AP}\|=\sqrt{(2-0)^2+(-1-y)^2+(1-0)^2}=\sqrt{5+(1+y)^2}$
The distance between $\displaystyle P(0,\,y,\,0)$ and $\displaystyle B(0,\,1,\,3)$ is $\displaystyle \|\vec{BP}\|=\sqrt{(0-0)^2+(1-y)^2+(3-0)^2}=\sqrt{9+(1-y)^2}$
We want these two distances to be equal thus $\displaystyle \|\vec{AP}\|=\|\vec{BP}\| \Leftrightarrow \sqrt{9+(1-y)^2}=\sqrt{5+(1+y)^2}$
Squaring both sides gives $\displaystyle 9+(1-y)^2=5+(1+y)^2$ which you can solve for $\displaystyle y$.
Does it help ?