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 : (1) `2x""+""6y""+""12 z""=""13` (2) `x""+""3y""+""6z""=-7` (3) `x""+""3y""+""6z""=""7` (4) `2x""+""6y""+""12 z""=-13`

Options

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

Answer

A

Explanation

The equation of the plane containing the line `2x-5y+z-3=0=x+y+4z-5` is
`(2x-5y+z-3)+lamda(x+y+4z-5)=0`
or`x(lamda+2)+y(lamda-5)+z(4lamda+1)-(5lamda+3)=0`……………..i
This is parallel to the plane `x+3y+6z-1=0`
`:.(lamda+2)/1=(lamda-5)/3=(4lamda+1)/6implieslamda=-11/2`
Puttng `lamda=-11/2` in i we obtain `x+3y+6z=y` as the equation of the required plane.