Question

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

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

Answer 1

The correct option is C

x + 3y + 6z = 7

Let equation of plane containing the lines.
$2x-5y+z=3andx+y+4z=5be(2x-5y+z-3)+\lambda (x+y+4z-5)=0\Rightarrow (2+\lambda )x+(\lambda -5)y+(4\lambda +1)z-3-5\lambda =0$
This plane is parallel to the plane x + 3y + 6z =1.
$\therefore \frac{2+\lambda }{1}=\frac{\lambda -5}{3}=\frac{4\lambda +1}{6}$
On taking first two equations, we get
$6+3\lambda =\lambda -5\Rightarrow 2\lambda =-11\Rightarrow \lambda =-\frac{11}{2}$
On taking last two equations, we get
$6\lambda -30=3+12\lambda \Rightarrow -6\lambda =33\Rightarrow \lambda =-\frac{11}{2}$
So, the equation of required plane is
$(2-\frac{11}{2})x+(\frac{-11}{2}-5)y+(-\frac{44}{2}+1)z-3+5\times \frac{11}{2}=0\Rightarrow -\frac{7}{2}x-\frac{21}{2}y-\frac{42}{2}z+\frac{49}{2}=0\Rightarrow x+3y+6z-7=0$