Find the equations of the planes parallel to the plane `x-2y+2z-3=0` which is at a unit distance from the point `(1,2,3)dot`

Answer

`x-2y+2z=0` or `x-2y+2z-6=0`

Explanation

Let the required plane be `x-2y+2z+k=0` for some constants k.
Then, its distance from the point P(1,2,3) is
`(|1-2xx2+2xx3+k|)/sqrt(1^(2)+(-2)^(2)+2^(2))=|3+k|=3`.
`rArr 3+k=3` or `3+k=-3 rArr k=0` or `k=-6`.
Hence, the required equations are `x-2y+2z=0` or `x-2y+2z-6=0`.