Answer:
Option C
Explanation:
Let equation of plane containing the line 2x-5y+z=3 and x+y+4z=5 be
$(2x-5y+z-3)+\lambda (x+y+4z-5)=0$
$\Rightarrow (2+\lambda)x+(\lambda-5)y+(4\lambda+1)z-3-5\lambda=0$ ........(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 wo equalities , we get
$6+3\lambda=\lambda-5$
$\Rightarrow$ $2\lambda=-11$
$\Rightarrow$ $\lambda=-\frac{11}{2}$
On taking last 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
$\left(2-\frac{11}{2}\right)x+\left(\frac{-11}{2}-5\right) y+\left(-\frac{44}{2}+1\right) z-3+5\times \frac{11}{2}=0$
$\Rightarrow$ $-\frac{7}{2}x-\frac{21}{2}y-\frac{42}{2}z+\frac{49}{2}=0$
$\Rightarrow$ x+3y+6z-7=0
