Equation of plane passing containing the intersection of other two planes, with a fixed distance to the origin. Hello all this is my first math question, I hope this is the right place to post.
Anyway, this is basic stuff but I'm a bit confused here, here's the problem's legend:
"Find the equation of the plane that passes through the intersection line ($l$) of the planes $2x-y-z=2, x+y-3z+4=0$ and has a distance of 2 from the origin.
Ok, So here's what I've done, the family of planes that pases through $l$ is $$2x-y-z-2+k(x+y-3z+4)=0$$ We work the equation so it looks more like a planes general form: $$Ax+By+Cz+D=0$$ giving us this:$$(2+k)x+(k-1)y+(-3k-1)z +2(2k-1)=$$
where$$A=(2+k)$$ $$B=(k-1)$$ $$C=(-1-3k)$$ $$D=2(2k-1)$$
The problem states that it's distance to the origin is 2 and the normal form of a plane($x\cos{\alpha}+y\cos{\beta}+z\cos{\gamma}-p=0$) where $p$ is the distance of the plane to the origin. We know that to transform a planes equation from the general form to the normal form we must divide each term by $$r=\pm \sqrt{A^2+B^2+C^2}$$
So $p=\frac{D}{r}$ and we have this:$$2=\frac{2(2k-1)}{\pm \sqrt{(2+k)^2+(k-1)^2+(-1-3k)^2}}$$
And here is where my doubt is, what sign should $r$ have positive or negative, the following theorem says:

*if $D \neq 0$, $r$ is opposite sign of $D$

*if $D=0$ and $C \neq 0$ $r$ is equal sign of $C$

*if $D=0$ and $C=0$ and $B \neq 0$ $r$ is equal sign of $B$

*if $D=0$ and $C=0$ and $B=0$ and $A \neq 0$ $r$ is equal sign of $A$
Since we don't know the value of k and $D(k)=2(2k-1)$ and to have $D$ we must find k but k is determined by the sign of D(what a mess hehe) how do we chose $r$'s sign?
Any help is really appreciated.
 A: Correct me if I'm wrong here, but if two planes are not co-planar (and these are not), then the intersection between the plane will form a single line.  That line will be whatever distance it happens to be from the origin.  I'm assuming that by distance from origin, they mean closest point from origin to the line.  But anyways, that distance from the line seems just extra.
Then again, I'm probably missing the point here.  All I can think of when seeing this problem is forming normals for the 2 planes and forming the cross product of those vectors to get the intersecting vector.  Been doing computer graphics for too long :)
A: First of all, your equation should be $$(k+1)x+(k-1)y-\cdots.$$
Note that the distance formula is  $$2=\frac{|2(2k-1)|}{\cdots}$$ (no $\pm$). Square both sides (note that square of $|\text{anything}|$ is $\text{anything}^2$, then solve a quadratic equation. You will get two $k$'s. Substitute back to see which one is correct. In this case, it turns out that both $k$'s are correct.
A: The adoption the + or – is essential in writing the equations of resultant planes (there are two in total) explicitly when other situations with different conditions given. The decision of using which sign is exactly like what you have mentioned. 
However, in this case, the sign is implicitly controlled by ‘k’ and the required equations ‘come out automatically without the necessity of considering the + / – sign’. See below.
Following your (and also corrections from others) work, we have
$ 1=\frac{|(2k-1)|}{|±\sqrt{A^2+B^2+C^2}|}$
$|±\sqrt{A^2+B^2+C^2}| = |(2k-1)|$
$A^2+B^2+C^2 = (2k-1)^2$
i.e. $1 + 2k + k^2 + k^2 – 2k + 1 + 9 + 6k + k^2 = 4k^2 – 4k + 1$
:
$k = 10.9xxx$ or $–0.9xxxxx$
Plug these values in correspondingly, you will get 2 sets of A, B, C, D and hence two equations for the two resultant planes meeting your requirement. This is the beauty of analytic geometry.
The following is an 2-D analogue of the said situation:-
