Show that the point `(1,2,1)` is equidistant from the planes `vecr.(hati+2hatj-2hatk) = 5` and `vecr.(2hati-2hatj+hatk) + 3 = 0`.

Explanation

Perpendicular distance from point `(1,2,1)` to the plane `vecr.(hati+2hatj-2hatk)-5 = 0`
`=|((hati+2hatj+hatk).(hati+2hatj-2hatk))/(sqrt(1+4+4))|`
`=|(1+4-2-5)/(3)|=2/3` units
Similarly the perpendicular distance from poin t `(1,2,1)` to the plane ` vecr.(2hati-2hatj+hatk) + 3 = 0`
`=|((hati+2hatj+hatk).(2hati-2hatj+hatk))/(sqrt(4+4+1))|`
`= |(2-4+1+3)/(3)|=2/3` units
Therefore point (1,2,1) is equidistant from the given two planes.