The point in YZ-plane which is equidistant from three pointsA(2, 0, 3), B(0, 3, 2) and C(0, 0, 1) is 

Since x-coordinate of every point in YZ-plane is zero. Let P(O, y, z) be a point on the YZ-plane such that PA=PB=PC.

 Now ,PA=PB

$\begin{array}{l}\Rightarrow (0-2{)}^2+(y-0{)}^2+(z-3{)}^2\\ =(0-0{)}^2+(y-3{)}^2+(z-2{)}^2,\\ \text{ i.e., }z-3y=0\end{array}$

and  PB=PC 

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{y}}^2+9-6\mathrm{y}+{\mathrm{z}}^2+4-4\mathrm{z}={\mathrm{y}}^2+{\mathrm{z}}^2+1-2\mathrm{z}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ i.e., }3\mathrm{y}+\mathrm{z}=6\end{array}$

On simplifying the two equations, we get y= 1 and z= 3. Here, the coordinate of the point Pare (0, 1, 3).