Question:
If the equation of a plane $P$, passing through the intersection of the planes, $x+4 y-z+7=0$ and $3 x+y+5 z=8$ is $a x+b y$
Solution:
Equation of plane $P$ is
$(x+4 y-z+7)+\lambda(3 x+y+5 z-8)=0$
$\Rightarrow x(1+3 \lambda)+y(4+\lambda)+z(-1+5 \lambda)+(7-8 \lambda)=0$
$\Rightarrow \frac{1+3 \lambda}{a}=\frac{4+\lambda}{b}=\frac{5 \lambda-1}{6}=\frac{7-8 \lambda}{-15}$
From last two ratios, $\lambda=-1$
$\Rightarrow \frac{-2}{a}=\frac{3}{b}=-1$
$\therefore a=2, b=-3$
$\therefore$ Equation of plane is, $2 x-3 y+6 z-15=0$
Distance $=\frac{|6-6-6-15|}{7}=\frac{21}{7}=3$.