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

A plane passes through (1,-2,1) and is perpendicualr to two planes` 2x

1.	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
Answer» Correct Answer - D 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)`

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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

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