The equation of the plane passing through the intersection of the planes `2x-5y+z=3` and `x+y+4z=5` and parallel to the plane `x+3y+6z=1` is `x+3y+6z=k`, where k is

Options

A. 5
B. 3
C. 7
D. 2

Answer

C

Explanation

Equation of plane passing through the intersection of the planes `2x-5y+z and x+y+4z=5` is
`(2x-5y+z-3)+lamda(x+y+4z-5)=0`
`rArr(2+lamda)x+(-5+lamda)y+1+4lamda)-3-5lamda=0` . . . (i)
which is parallel to the plane `x+3y+6z=1`.
Therefore, `(2+lamda)/(1)=(-5+lamda)/(3)=(1+4lamda)/(6)rArrlamda=(-11)/(2)`
From Eq. (i), `(2-(11)/(2))x+(-5-(11)/(2))y+(1+4xx((-11)/(2)))z-3-5xx(-(11)/(2))=0`
`rArr" "-(7)/(2)x-(21)/(2)y-21z+(49)/(2)=0rArrx+3y+6z=7`
Hence, `k=7`