Find the point in xy-plane which is equidistant from the points

The general point on xy plane is D(x, y, 0). Consider this point is equidistant to the points A(2, 0, 3), B(0, 3, 2) and C(0, 0, 1).

∴ AD = BD

$\sqrt{(x-2)^{2}+(y-0)^{2}+(0-3)^{2}}=\sqrt{(x-0)^{2}+(y-3)^{2}+(0-2)^{2}}$

Squaring both sides

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

$x^{2}-4 x+4+y^{2}+9=x^{2}+y^{2}-6 y+9+4$

$-4 x=-6 y \ldots(1)$

Also, AD = CD

$\sqrt{(x-2)^{2}+(y-0)^{2}+(0-3)^{2}}=\sqrt{(x-0)^{2}+(y-0)^{2}+(0-1)^{2}}$

Squaring both sides,

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

$x^{2}-4 x+4+y^{2}+9=x^{2}+y^{2}+1$

$-4 x=-12$ ..........(2)

Simultaneously solving equation (1) and (2) we get

X = 3, y = 2.

The point which is equidistant to the points A(2, 0, 3), B(0, 3, 2) and C(0, 0, 1) is (3, 2, 0).