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. `sqrt2`
D. `2sqrt2`

Answer

d

Explanation

The equation of the plane passing through the point `(1, -2, 1)` and perpendicular to the planes `2x-2y+z=0 and x-y+2z=4` is given by
`|{:(x-1,,y+2,,z-1),(2,,-2,,1),(1,,-1,,2):}|=0`
or `" "x+y+1=0`
Its distance from the point `(1, 2, 2)` is
`" "|(1+2+1)/(sqrt2)|= 2sqrt2`