The equation of a plane passing through the intersection of two planes $x+2 y-3 z+2=0$ and $6 x+y+z+1=0$ and…
The equation of a plane passing through the intersection of two planes $x+2 y-3 z+2=0$ and $6 x+y+z+1=0$ and parallel to the line $x-1=y+2=7-z$ is
$5 x-y+4 z+1=0$
$5 x+y+4 z+1=0$
$5 x-y+4 z=1$
$5 x+y+4 z=1$
Solution
Equation of plane passing through the line of intersection of given planes is $(x+2 y-3 z+2)+\lambda(6 x+y+z+1)=0$
$(1+6 \lambda) x+(2+\lambda) y+(-3+\lambda) z+(2+\lambda)=0$
This is parallel to the line $\frac{x-1}{1}=\frac{y+2}{1}=\frac{z-7}{-1}$
$\therefore(1+6 \lambda)(1)+(2+\lambda)(1)+(-3+\lambda)(-1)=0 \Rightarrow \lambda=-1$
Hence required equation of plane is
$-5 x+y-4 z+1=0 \Rightarrow 5 x-y+4 z=1$