Question

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

Answer: (C)

$x+3y+6z=7$

Let equation of plane containing the lines

$2x-5y+z=3$ and $x+y+4z=5$ be

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

or, $\left(2+\lambda \right)x+\left(\lambda -5\right)y+\left(4\lambda +1\right)z-3-5\lambda =\mathrm{0........}\left(i\right)$

This planes parallel to the plane $x+3y+6z=1$

$\therefore \frac{2+\lambda }{1}=\frac{\lambda -5}{3}=\frac{4\lambda +1}{6}$

Considering first two qualities, we get

$\begin{array}{c}\Rightarrow 6+3\lambda =\lambda -5\\ \Rightarrow 2\lambda =-11\\ \Rightarrow \lambda =\frac{-11}{2}\end{array}$

Considering the last two qualities ,we get

$\begin{array}{c}\Rightarrow 6\lambda -30=3+12\lambda \\ \Rightarrow -6\lambda =33\\ \Rightarrow \lambda =\frac{-11}{2}\end{array}$

So, the equation of required plane is

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

$\begin{array}{c}\Rightarrow \frac{-7}{2}x-\frac{21}{2}y-\frac{42}{2}z+\frac{49}{2}=0\\ \Rightarrow x+3y+6z-7=0\\ \Rightarrow x+3y+6z=7\end{array}$

Hence,

Option $C$ is correct answer