Question

If the equation of a plane $P$, passing through the intersection of the planes, $x+4y-z+7=0$ and $3x+y+5z=8$ is $ax+by+6z=15$ for some $a,b\in R,$ then the distance of the point $(3,2,-1)$ from the plane $P$ is

Answer 1

$p_1+\lambda p_2=0$
$\Rightarrow (x+4y-z+7)+\lambda (3x+y+5z-8)$
$\Rightarrow x(1+3\lambda )+y(4+\lambda )+z(5\lambda -1)+(7-8\lambda )=0$
Compare with the equation of plane $ax+by+6z-15=0$

$\frac{1+3\lambda }{a}=\frac{4+\lambda }{b}=\frac{-1+5\lambda }{6}=\frac{7-8\lambda }{-15}$
$\therefore 15-75\lambda =42-48\lambda$
$\Rightarrow -27=27\lambda$
$\Rightarrow \lambda =-1$
$\therefore P:-2x+3y-6z+15=0$
Now, the distance of the point $(3,2,-1)$ from the plane $P$ is
$d=\left|\frac{-6+6+6+15}{\sqrt{4+9+36}}\right|=3$ units