Find the distance of the point (2,3,-5) from the plane `x+2y-2z=9`.

Explanation

Required distance.
=Length of perpendicular from (2,3,-5) to the plane `x+2y-2z-9=0`.
`=(|(2+2 xx 3-2xx(-5)-9)|)/sqrt(1^(2)+2^(2)+(-2)^(2))=(|2+6+10-9|)/sqrt(9)`
`=9/3=3` units.
Hence, the required distance is 3 units.