Find the coordinates of the point which is equidistant from the points

Consider, D(x,y,z) point equidistant from points A(a, 0, 0), B(0, b, 0), C(0, 0, c) and O(0, 0, 0).

∴ AD = 0D

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

Squaring both sides,

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

$x^{2}+2 a x+a^{2}+y^{2}+z^{2}=x^{2}+y^{2}+z^{2}$

$a(2 x-a)=0$

as $a \neq 0$

$x=a / 2$

∴ BD = 0D

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

Squaring both sides,

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

$x^{2}+y^{2}+2 b y+b^{2}+z^{2}=x^{2}+y^{2}+z^{2}$

$b(2 y-b)=0$

as $b \neq 0$

$y=b / 2$

$\therefore C D=0 D$

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

Squaring both sides,

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

$x^{2}+y^{2}+z^{2}+2 c z+c^{2}=x^{2}+y^{2}+z^{2}$

$c(2 z-c)=0$

as $c \neq 0 .$

$z=c / 2$

Therefore, the pint $\mathrm{D}(\mathrm{a} / 2, \mathrm{~b} / 2, \mathrm{c} / 2)$ is equidistant to points $\mathrm{A}(\mathrm{a}, 0,0), \mathrm{B}(0, \mathrm{~b}, 0), \mathrm{C}(0,0, \mathrm{c})$ and $O(0,0,0)$.