The equation of the plane containing the line `2x-5y""+""z""=""3;""x""+""y""+""4z""=""5` , and parallel to the plane, `x""+""3y""+""6z""=""1` , is :

Options

A. `2x+6y+12z=13`
B. `x+3y+6z=-7`
C. `x+3y+6z=7`
D. `2x+6y+12z=-13`

Answer

C

Explanation

Let equation of plane containing the lines
`2x-5y+z=3andx+y+4z=5" be"`
`(2x-5y+z-3)+lamda(x+y+4z-5)=0`
`rArr(2+lamda)x+(lamda-5)y+(4lamda+1)z-3-5lamda=0` . . . (i)
This plane is parallel to the plane `x+3y+6z=1`.
`:." "(2+lamda)/(1)=(lamda-5)/(3)=(4lamda+1)/(6)`
On taking first two equalities, we get
`6+3lamda=lamda-5`
`rArr" "lamda=-(11)/(2)`
On taking last two equalities, we get
`6lamda-30=3+12lamda`
`rArr" "lamda=-(11)/(2)`
So, the equation of required plane is
`(2-(11)/(2))x+((-11)/(2)-5)y+(-(44)/(2)+1)z-3+5xx(11)/(2)=0`
`rArr" "-(7)/(2)x-(21)/(2)y-(42)/(2)z+(49)/(2)=0`
`rArr" "x+3y+6z-7=0`