A plane which is perpendicular to two planes `2x-2y+z=0 and x-y+2z = 4` passes through `(1, -2, 1)`. The distance of the plane from the point (1, 2, 2) is

Options

A. 0
B. 1
C. `sqrt(2)`
D. `2sqrt(2)`

Answer

D

Explanation

The equation of a plane passing through (1,-2,1) is
`a(x-1)+b(y+2)+c(z-1)=0`……………i
It is perpendicular to `2x-2y+z=0` and `x-y+2z=4`
`:.2a-2b+c=0` and `a-b+2c=0`
`impliesa/(-3)=b/(-3)=c/0`
Substituting the values of `a,b,c ` in(i) we obtain
`x+y+1=0`
Distance of this plane from `(1,2,2)` is given by
`d=(1+2+1)/(sqrt(1+1))=2sqrt(2)`