Question

Find the equation of planes which passes through the intersection of the lines $x+3y+6=0$ and $3x-y-4z=0$ and their perpendicular distance from the origin is $1$.

Answer 1

Equation of the line passing through the intersection of the given plane is given by
$(x+3y+6)+k(3y-y-4z)=0$
$(1+3k)x+(3-k)y-4kz+6=0$
If the perpendicular to it from $(0,0,0)$ is $p$, then
$p=\left|\frac{0+0-0+6}{\sqrt{{(1+3k)}^2+{(3-k)}^2+{(-4k)}^2}}\right|$
$36={(1+3k)}^2+{(3-k)}^2+16k^2$
Solving $k=\pm 1$
The required planes are
$(x+3y+6)\pm (3x-y-4z)=0$
$4x+2y-4z+6=0$
and $-2x+4y+4z+6=0$