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. $2x+6y+12z=-13$

Answer 1

The correct option is C $x+3y+6z=7$

Let equation of plane containing the lines $2x-5y+z=3$ and $x+y+4z=5$ be

$(2x-5y+z-3)+\lambda (x+y+4z-5)=0$

$\Rightarrow (2+\lambda )x+(\lambda -5)y+(4\lambda +1)z-3-5\lambda =0\dots$ (i)

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 equalities, we get

$6\lambda -30=3+12\lambda$

$\Rightarrow -6\lambda =33$

$\Rightarrow \lambda =\frac{-11}{2}$

So, the equation of required plane is

$(2-\frac{11}{2})x+(\frac{-11}{5}-5)y+(\frac{-44}{2}+1)z-3+5\times \frac{11}{2}=0$

$\Rightarrow -\frac{7}{2}x-\frac{21}{2}y-\frac{42}{2}z+\frac{49}{2}=0$

$\Rightarrow x+3y+6z-7=0$