Show that the distance between planes `2x-2y+z+3=0 and 4x-4y+2z+5=0 is 1/6`

Explanation

Equation of first plane
`3x+2y-z + 1 = 0`
Let `(x_(1),y_(1),z_(1))` be any point on this plane.
`:. 3x_(1)+2y_(1)-z_(1) + 1 = 0 "……"(1)`
Perpendcicular distance from point `(x_(1),y_(1)z_(1))` to the second plane `3x+2y-z+5 = 0`
`=|(3x_(1)+2y_(1)-z_(1)+5)/(sqrt(3^(2)+2^(2)+(-1)^(2)))|` , [From eq.(1)]
`= |((-1)+5)/(sqrt(9+4+1))|=4/(sqrt(14))` units
Therefore the distance between parallel planes `= (4)/(sqrt(14))` units.