The equation of the plane through the point (2,5,-3) and perpendicular to the planes `x+2y+2z=1` and `x-2y+3z=4` is

Options

A. `3x-4y+2z-2=0`
B. `7x-y+5z=30`
C. `x-2y+z=11`
D. `10x-y-4z=27`

Answer

D

Explanation

Equation of plane through (2,5,-3) is
`a(x-2)+b(y-5)+c(z+3)=0` . . . (i)
which is perpendicular to
`x+2y+2z=1andx-2y+3z=4`
then `a+2b+2c=0` . . . (ii)
and `a-2b+3c=0` . . . (iii)
Eliminating a,b,c from Eqs. (i), (ii) and (iii), we get
`|{:(x-2,y-5,z+3),(1,-2,2),(1,-2,3):}|=0`
`rArr(x-2)(10)-(y-5)(1)+(z+3)(-4)=0`
`rArr" "10x-y4z-27=0`