A [sphere](https://artofproblemsolving.com/wiki/index.php/Sphere) is inscribed in the [tetrahedron](https://artofproblemsolving.com/wiki/index.php/Tetrahedron) whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$  The [radius](https://artofproblemsolving.com/wiki/index.php/Radius) of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m + n.$
The center $I$ of the insphere must be located at $(r,r,r)$ where $r$ is the sphere's radius.
$I$ must also be a distance $r$ from the plane $ABC$
The signed distance between a plane and a point $I$ can be calculated as $\frac{(I-G) \cdot P}{|P|}$, where G is any point on the plane, and P is a vector perpendicular to ABC.
A vector $P$ perpendicular to plane $ABC$ can be found as $V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle$
Thus $\frac{(I-C) \cdot P}{|P|}=-r$ where the negative comes from the fact that we want $I$ to be in the opposite direction of $P$
\begin{align*}\frac{(I-C) \cdot P}{|P|}&=-r\\ \frac{(\langle r, r, r \rangle-\langle 0, 0, 2 \rangle) \cdot P}{|P|}&=-r\\ \frac{\langle r, r, r-2 \rangle \cdot \langle 8, 12, 24 \rangle}{\langle 8, 12, 24 \rangle}&=-r\\ \frac{44r -48}{28}&=-r\\ 44r-48&=-28r\\ 72r&=48\\ r&=\frac{2}{3} \end{align*}

Finally $2+3=\boxed{005}$