The point in $XY-plane$which is equidistant from three points 

$A(2,0,3), B(0,3,2)$ and $C(0,0,1)$ has the coordinates 

We know that $z$-coordinate of every point on $xy$-plane  is zero.

So, let $P(x,y,0)$ be a point on $xy$- plane such that $PA=PB=PC$

Now,  $PA= PB$

$\begin{array}{l}\Rightarrow P{A}^2=P{B}^2\\ \Rightarrow (x-2{)}^2+(y-0{)}^2+(0-3{)}^2=(x-0{)}^2+(y-3{)}^2+(0-2{)}^2\\ \Rightarrow 4x-6y=0\Rightarrow 2x-3y=0 ...\left(i\right)\end{array}$

and,

$\begin{array}{l} PB=PC\\ \Rightarrow P{B}^2=P{C}^2\\ \Rightarrow (x-0{)}^2+(y-3{)}^2+(0-2{)}^2=(x-0{)}^2+(y-0{)}^2+(0-1{)}^2\\ \Rightarrow -6y+12=0\\ \Rightarrow y=2 ...\left(ii\right)\end{array}$

Putting $y = 2$ in $\left(i\right)$, we obtain$x = 3$

Hence, the required point is $(3,2,0)$