Find the coordinates of the point which is equidistant from the four points `O(0,0,-0),\ A(2,0,0),\ B(0,3,0)a n d\ C(0,0,8)dot`

Explanation

Given points are `O(0,0,-0),\ A(2,0,0),\ B(0,3,0)a n d\ C(0,0,8)`
we have to Find the coordinates of the point which is equidistant from the four
points.
Let required point P(x, y, z).
so, according to question `PA = PB = PC = PO`
`=>PA^2=PB^2=PC^2=PO^2`
The distance between P and O is
`PO=sqrt((x-0)^2+(y-0)^2+(z+0)^2`
`=sqrt(x^2+y^2+z^2)`
The distance between P and A is
`PA=sqrt((x-2)^2+(y-0)^2+(z-0)^2`
`=sqrt((x-2)^2+y^2+z^2)`
The distance between P and B is
`PB=sqrt((x-0)^2+(y-3)^2+(z-0)^2`
`=sqrt(x^2+(y-3)^2+z^2)`
The distance between P and C is
`PC=sqrt((x-0)^2+(y-0)^2+(z-8)^2`
`=sqrt(x^2+y^2+(z-8)^2)`
as `PO^2=PA^2`
`x^2+y^2+z^2=(x-2)^2+y^2+z^2`
`=>x^2=x^2+4-4x`
`=>x=1`
again `PO^2=PB^2`
`x^2+y^2+z^2=(x^2+(y-3)^2+z^2`
`=>x^2+y^2=x^2+y^2+9-6y`
`=>y=3/2`
again `PO^2=PC^2`
`x^2+y^2+z^2=x^2+y^2+(z-8)^2`
`=>z=4`
Hence the point is `(1,3/2,4)`.