Determine the point in `X Y-p l a n e` which is equidistant from three points `A(2,0,3),B(0,3,2)` and `C(0,0,1)dot`

Options

A. `(2,0,8)`
B. `(0,3,1)`
C. `(3,2,0)`
D. `(3,2,1)`

Answer

C

Explanation

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`
`impliesPA^(2)=PB^(2)`
`implies(x-2)^(2)+(y-0)^(2)+(0-3)^(2)=(x-0)^(2)+(y-3)^(2)+(0-2)^(2)`
`implies4x-6y=0implies2x-3y=0`..............`(i)`
and
`PB=PC`
`impliesPB^(2)=PC^(2)`
`implies(x-0)^(2)+(y-3)^(2)+(0-2)^(2)=(x-0)^(2)+(y-0)^(2)+(0-1)^(2)`
`implies-6y+12=0`
`impliesy=2` .............`(ii)`
Putting `y=2` in `(i)` we obtain `x=3`
Hence the required point in `(3,2,0)`.