How to find the equation of the plane passing through the intersection of two other planes and whose perpendicular distance from the origin is given? 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.
 A: Any plane passing through the intersection of the given planes can be written in the form $x+3y+6 + k(3x-y-4z) = 0$. The perpendicular distance of this from the origin is
$$\frac{6}{\sqrt{(1+3k)^2+(3-k)^2+16k^2}} = \pm 1$$
and hence we have 
$36k^2 = 36$ and $k = \pm 1$. Thus the planes are $2x+y-2z+3 = 0$ and $-x+2y+2z+6=0$
A: When  the equation of the plane is written in the form $\vec n \cdot \vec r = d$, where $\vec n$ is the unit normal vector to the plane, directed away from the origin, and $\vec r$ is a point on the plane, then $d$ is the distance from the plane to the origin. 
So in this case, the equation is $\vec n \cdot \vec r = 1$. But note the requirement that $\vec n$ be a unit vector. So one equation for $\vec n$ is that $\|\vec n\|^2 = 1$. 
Since the line of intersection lies within our plane, the direction vector for that line has to point along the plane, and therefore must be normal to $\vec n$. So a second equation for $\vec n$ is $\vec n \cdot \vec v = 0$, where $\vec v = (-12, 4,-10)$. 
Finally, there are numerous lines with this same direction vector. But there is only one that we are interested in: the line that is the intersection of the other two planes. To get this line and not any of the others, we need a point on this line. This is fairly easy to find. Pick a value for one of the variables, then solve the system of the two given planar equations to find the other two variables. Once you have this point, call it $\vec r$, recall that since it is a point on the plane you are looking for, it satisfies the planar equation. Which gives you a third equation in $\vec n$:
$$\vec n \cdot \vec r = 1$$
$$\|\vec n \|^2 = 1$$
$$\vec n \cdot \vec v = 0$$
This is three equations in the three unknowns of the components of $\vec n$. So solve for them and you are done.
