I think I see how it works. At least, I didn't understand the explination, but I see where A' comes in. (As I am too lazy to find the fancy characters, I will use A for the plane and a for the line.)
Given: line a
plane B || line a
line a is within plane A
not(plane A || plane B) | already proven that this can happen
Prove: a || the intersection of A and B
Let there be a plane A', which is parallel to B and contains a.
Since A and B intersect, there must be a line b that is contained by both A and B.
So now there are parallel planes A' and B, each containing a line (a and b, repectively).
Since a and b are both within plane A, they must either intersect or be parallel. However, they are also the intersections of A and B and A and A', and B and A' are defined to be parallel, a and b must also be parallel.
Moreover, since b is defined as the intersection of A and B, a is parallel the intersection of A and B. And there is your proof.
Sorry if that was sketchy. I don't remeber my theorums from last year.