Answer: 5

Solution:

Connecting the incenter to the four vertices splits the tetrahedron into four pyramids of height $r$ over the faces, so $V = \frac{1}{3} r S,$ i.e. $r = \frac{3V}{S},$ where $S$ is the total surface area.

Here $V = \frac{1}{6} \cdot 6 \cdot 4 \cdot 2 = 8.$ The three faces on the coordinate planes are right triangles with areas $\frac{1}{2} \cdot 6 \cdot 4 = 12,$ $\frac{1}{2} \cdot 6 \cdot 2 = 6,$ and $\frac{1}{2} \cdot 4 \cdot 2 = 4.$ For face $ABC,$ the cross product of $\overrightarrow{AB} = (-6, 4, 0)$ and $\overrightarrow{AC} = (-6, 0, 2)$ is $(8, 12, 24),$ with length $4\sqrt{2^2 + 3^2 + 6^2} = 28,$ so that face has area $14.$

Then $S = 12 + 6 + 4 + 14 = 36$ and $r = \frac{24}{36} = \frac{2}{3},$ giving $m + n = 2 + 3 = 5.$