Let us call the point where $\overline{EF}$ intersects $\overline{AD}$ point $G$, and the point where $\overline{EF}$ intersects $\overline{BC}$ point $H$. Since angles $FHB$ and $EGA$ are both right angles, and angles $BEF$ and $DFE$ are congruent due to parallelism, right triangles $BHE$ and $DGF$ are similar. This implies that $\frac{BH}{GD} = \frac{9}{8}$. Since $BC=10$, $BH+GD=BH+HC=BC=10$. ($HC$ is the same as $GD$ because they are opposite sides of a rectangle.) Now, we have a system:
$\frac{BH}{GD}=\frac{9}8$
$BH+GD=10$
Solving this system (easiest by substitution), we get that:
$BH=\frac{90}{17}$
$GD=\frac{80}{17}$
Using the Pythagorean Theorem, we can solve for the remaining sides of the two right triangles:
$\sqrt{9^2-\left(\frac{90}{17}\right)^2}$ and $\sqrt{8^2-\left(\frac{80}{17}\right)^2}$
Notice that adding these two sides would give us twelve plus the overlap $EF$. This means that:
$EF= \sqrt{9^2-\left(\frac{90}{17}\right)^2}+\sqrt{8^2-\left(\frac{80}{17}\right)^2}-12=3\sqrt{21}-12$
Since $21$ isn't divisible by any perfect square, our answer is:
$3+21+12=\boxed{36}$
In rectangle $ABCD$, $AB = 12$ and $BC = 10$.  Points $E$ and $F$ lie inside rectangle $ABCD$ so that $BE = 9$, $DF = 8$, $\overline{BE} \parallel \overline{DF}$, $\overline{EF} \parallel \overline{AB}$, and line $BE$ intersects segment $\overline{AD}$.  The length $EF$ can be expressed in the form $m \sqrt{n} - p$, where $m$, $n$, and $p$ are positive integers and $n$ is not divisible by the square of any prime.  Find $m + n + p$.