Question

The equation of the plane containing the lines $2x-5y+z=3,$ $x+y+4z=5$ 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 $x+3y+6z=7$

Given : $p_1:2x-5y+z-3=0$ and $p_2:x+y+4z-5=0$
Any plane through the line of intersection is $p_1+\lambda p_2=0$
The equation of required plane can be obtained as:
$2x-5y+z-3+\lambda (x+y+4z-5)=0\cdots \left(i\right)$
As plane is parallel to $x+3y+6z-1=0,$
$\frac{2+\lambda }{1}=\frac{\lambda -5}{3}=\frac{1+4\lambda }{6}$
$\Rightarrow 6+3\lambda =\lambda -5\Rightarrow \lambda =-\frac{11}{2}$
Substituting the value of $\lambda$ in $\left(i\right)$, we have
$-7x-21y-42z+49=0$
$\Rightarrow x+3y+6z-7=0$