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

(3, 2, 0)

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

$⇒ PA^2 = PB^2 $

$⇒ (x-2)^2 + (y-0)^2 + (0-3)^2 = (x-0)^2 +(y-3)^2 + (0-2)^2 $

$⇒ 4x - 6y = 0 ⇒ 2x - 3y = 0 $ ....................(i)

and,

$PB = PC$

$⇒ PB^2 = PC^2 $

$⇒ (x-0)^2 + (y-3)^2 + (0-2)^2 = (x-0)^2 +(y-0)^2 + (0-1)^2 $

$⇒ -6y + 12 = 0 $

$⇒ y = 2 $ ...........(ii)

Putting y = 2 in (i), we obtain x = 3

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