In triangle $ABC$, $BC = 23$, $CA = 27$, and $AB = 30$. Points $V$ and $W$ are on $\overline{AC}$ with $V$ on $\overline{AW}$, points $X$ and $Y$ are on $\overline{BC}$ with $X$ on $\overline{CY}$, and points $Z$ and $U$ are on $\overline{AB}$ with $Z$ on $\overline{BU}$. In addition, the points are positioned so that $\overline{UV}\parallel\overline{BC}$, $\overline{WX}\parallel\overline{AB}$, and $\overline{YZ}\parallel\overline{CA}$. Right angle folds are then made along $\overline{UV}$, $\overline{WX}$, and $\overline{YZ}$. The resulting figure is placed on a leveled floor to make a table with triangular legs. Let $h$ be the maximum possible height of a table constructed from triangle $ABC$ whose top is parallel to the floor. Then $h$ can be written in the form $\frac{k\sqrt{m}}{n}$, where $k$ and $n$ are relatively prime positive integers and $m$ is a positive integer that is not divisible by the square of any prime. Find $k+m+n$.
Note that triangles $\triangle AUV, \triangle BYZ$ and $\triangle CWX$ all have the same height because when they are folded up to create the legs of the table, the top needs to be parallel to the floor. We want to find the maximum possible value of this height, given that no two of $\overline{UV}, \overline{WX}$ and $\overline{YZ}$ intersect inside $\triangle ABC$. Let $h_{A}$ denote the length of the altitude dropped from vertex $A,$ and define $h_{B}$ and $h_{C}$ similarly. Also let $\{u, v, w, x, y, z\} = \{AU, AV, CW, CX, BY, BZ\}$. Then by similar triangles
\begin{align} \frac{u}{AB}=\frac{v}{AC}=\frac{h}{h_{A}}, \\ \frac{w}{CA}=\frac{x}{CB}=\frac{h}{h_{C}}, \\ \frac{y}{BC}=\frac{z}{BA}=\frac{h}{h_{B}}. \end{align}
Since $h_{A}=\frac{2K}{23}$ and similarly for $27$ and $30,$ where $K$ is the area of $\triangle ABC,$ we can write
\begin{align} \frac{u}{30}=\frac{v}{27}=\frac{h}{\tfrac{2K}{23}}, \\ \frac{w}{27}=\frac{x}{23}=\frac{h}{\tfrac{2K}{30}}, \\ \frac{y}{23}=\frac{z}{30}=\frac{h}{\tfrac{2K}{27}}. \end{align}
and simplifying gives $u=x=\frac{690h}{2K}, v=y=\frac{621h}{2K}, w=z=\frac{810h}{2K}$. Because no two segments can intersect inside the triangle, we can form the inequalities $v+w\leq 27, x+y\leq 23,$ and $z+u\leq 30$. That is, all three of the inequalities
\begin{align} \frac{621h+810h}{2K}\leq 27, \\ \frac{690h+621h}{2K}\leq 23, \\ \frac{810h+690h}{2K}\leq 30. \end{align}
must hold. Dividing both sides of each equation by the RHS, we have
\begin{align} \frac{53h}{2K}\leq 1\, \text{since}\, \frac{1431}{27}=53, \\ \frac{57h}{2K}\leq 1\, \text{since}\, \frac{1311}{23}=57, \\ \frac{50h}{2K}\leq 1\, \text{since}\, \frac{1500}{30}=50. \end{align}
It is relatively easy to see that $\frac{57h}{2K}\leq 1$ restricts us the most since it cannot hold if the other two do not hold. The largest possible value of $h$ is thus $\frac{2K}{57},$ and note that by Heron's formula the area of $\triangle ABC$ is $20\sqrt{221}$. Then $\frac{2K}{57}=\frac{40\sqrt{221}}{57},$ and the answer is $40+221+57=261+57=\boxed{318}$
~sugar_rush