A plane passes through (1,-2,1) and is perpendicualr to two planes` 2x-2y+z=0" and "x-y+2z=4,` then 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

Let the equation of plane be
`a(x-1)+b(y+2)+c(z-1)=0`
which is perpendicular to `2x-2y+z=0` and `x-y+2z=4.`
`implies" "2a-2b+c=0" and "a-b+2c=0`
`implies" "(a)/(-3)=(b)/(-3)=(c)/(0)implies(a)/(1)=(b)/(1)=(c)/(0).`
So, the equation of plane is `x-1+y+2=0`
or`" "x+y+1=0`
Its distance from the point (1,2,2) is `(|1+2+1|)/(sqrt(2))=2sqrt(2)`