The equation of the plane passing through the point `(1,1,1)` and perpendicular to the planes `2x+y-2z=5` and `3x-6y-2z=7` is

Options

A. `14x+2y-15z=1`
B. `14x-2y+15z=27`
C. `14x+2y+15z=31`
D. `-14x+2y+15z=3`

Answer

C

Explanation

The equation of a plane through (1,1,) is
`a(x-1)+b(y-1)+c(z-1)=0`…………i
It is perpendicular to the planes `2x+y-2z=5` and `3x-6y-2z=7`
`:.2a+b-2c=0` and `3a-6b-b-2c=0`
`implies a/(-14)=b/(-2)=c/(-15)`
Substituting these values of a,b,c in i we obtain `14x+2y+15z=31` as the equation of the required plane.