I've been trying to find a way to work this out for hours now with no luck.
The question is:
Find the equation of the plane(s) passing through the intersection of the planes $x+3y+6=0$ and $3x-y-4z=0$ and whose perpendicular distance from the origin is unity.
What I've tried so far:
I found the direction of the line passing through the intersection by cross multiplying the normal vectors of the two given planes.
The direction of the line at the intersection is: $<-12, 4, -10>$
I'm thoroughly confused right now and have no idea what to do from here on. Finding the equation of the line at the intersection is something I can do, but I have no idea how to find the equation of a plane at the intersection which is also at a distance of $1$ from the origin. Any help would be much appreciated.