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

Equation of any plane which passes through the intersection of the planes 2x - 5y + z = 3 and x + y + 4z = 5 is given by
$2x-5y+z-3+\lambda (x+y+4z-5)=0$
$x(2+\lambda )+y(\lambda -5)+z(4\lambda +1)-3-5\lambda =0$
We are given this plane is parallel to x + 3y + 6z = 1. Comparing the coefficients, we get
$\Rightarrow \frac{\lambda +2}{1}=\frac{\lambda -5}{3}=\frac{4\lambda +1}{6}$
$3\lambda +6=\lambda -5$
$2\lambda =-11$
$\lambda =\frac{-11}{2}$
We will now substitute this value in the equation of plane.
$\Rightarrow equationofplane\frac{-7x}{2}-\frac{21y}{2}-21z+\frac{49}{2}=0\Rightarrow 7x+21y+42z-49=0$
$\Rightarrow x+3y+6z-7=0$