Find the distance between the parallel planes
`vecr.(2hati-3hatj+6hatk) = 5` and
`vecr.(6hati-9hatj+18hatk) + 20 = 0`.

Explanation

Taking `vecr=(xhati+yhatj+zhatk)_`, the Cartesian equations of the given planes are
`(xhati+yhatj+zhatk).(2hati-3hatj+6hatk)=5`
`rArr 2x-3y+6z-5=0`…………(i)
and `(xhati+yhatj+zhatk).(6hati-9hatj+18hatk)+20=0`
`rArr 6x-9y+18z+20=0`…………......(ii)
Let `P(x_(1),y_(1),z_(1))` be any point on (i). then, `2x_(1)-3y_(1)+6z_(1)=5`...............(iii)
Distance between the given planes.
=length of the perpendicular from `P(x_(1),y_(1),z_(1))` to the plane (ii) ,
`=(|6x_(1)-9y_(1)+18z_(1)+20|)/(sqrt(6^(2)+(-9)^(2)+18^(2))=(|3(2x_(1)-3y_(1)+6z_(1))+20|)/(sqrt(36+81+324)`
`=(|(3xx5)+20|)/(sqrt(441))` [using (iii)].
`=35/21=5/3` units.
Hence, the distance between the given planes is `5/3` units.