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 passing through (1,-2,1) be
`(x-1)+b(y+2)+c(z-1)=0`
which is perpendicular to `2x-2y+z=0`
`and" "x-y+2z=4`
`:." "2a-2b+c=0`
`[because a_(1)a_(2)+b_(1)b_(2)=0]`
`and" "a-b+2c=0`
`rArr" "(a)/(-3)=(b)/(-3)=(c)/(0)rArr(a)/(1)=(b)/(a)=(c)/(0)`
so, the equation of plane is
`(x-1)+(y+2)=0rArrx+y+1=0`, its distance from the points (1,2,2) is
`(|1+2+1|)/(sqrt(2))=2sqrt(2)`