When a light beam reflects off a surface, the path is like that of a ball bouncing. Picture that, and also imagine X, Y, and Z coordinates for the cube vertices. The coordinates will all involve 0's and 12's only, so that means that the X, Y, and Z distance traveled by the light must all be divisible by 12. Since the light's Y changes by 5 and the X changes by 7 (the Z changes by 12, don't worry about that), and 5 and 7 are relatively prime to 12, the light must make 12 reflections onto the XY plane or the face parallel to the XY plane. 
In each reflection, the distance traveled by the light is $\sqrt{ (12^2) + (5^2) + (7^2) }$ = $\sqrt{218}$. This happens 12 times, so the total distance is $12\sqrt{218}$. $m=12$ and $n=218$, so therefore, the answer is $m+n=\boxed{230}$.
Let $ABCD$ and $BCFG$ be two faces of a cube with $AB=12$. A beam of light emanates from vertex $A$ and reflects off face $BCFG$ at point $P$, which is 7 units from $\overline{BG}$ and 5 units from $\overline{BC}$. The beam continues to be reflected off the faces of the cube. The length of the light path from the time it leaves point $A$ until it next reaches a vertex of the cube is given by $m\sqrt{n}$, where $m$ and $n$ are integers and $n$ is not divisible by the square of any prime. Find $m+n$.