Three, (16 points) In acute $\triangle ABC$, it is known that $AB=$ 4, $AC=5, BC=6, AD, BE, CF$ are the altitudes to sides $BC, CA, AB$ respectively, with $D, E, F$ being the feet of the perpendiculars. Find $\frac{S_{\triangle DEF}}{S_{\triangle ABC}}$
Three, let $B D=x, C E=y$, $A F=z$, then by the Pythagorean theorem, we get
$$
\left\{\begin{array}{l}
4^{2}-x^{2}=5^{2}-(6-x)^{2}, \\
6^{2}-y^{2}=4^{2}-(5-y)^{2}, \\
5^{2}-z^{2}=6^{2}-(4-z)^{2} .
\end{array}\right.
$$

Solving, we get $x=\frac{9}{4}, y=\frac{9}{2}, z=\frac{5}{8}$.
$$
\begin{aligned}
\frac{S_{\triangle B D F}}{S_{\triangle A B C}}= & \frac{\frac{9}{4} \cdot \frac{27}{8}}{6 \cdot 4}=\frac{81}{256}, \\
\frac{S_{\triangle C D E}}{S_{\triangle A B C}} & =\frac{\frac{15}{4} \cdot \frac{9}{2}}{6 \cdot 5}=\frac{9}{16},
\end{aligned}
$$
$$
\begin{array}{l}
\frac{S_{\triangle A E F}}{S_{\triangle A B C}}=\frac{\frac{1}{2} \cdot \frac{5}{8}}{5 \cdot 4}=\frac{1}{64}, \\
\frac{S_{\triangle D E F}}{S_{\triangle A B C}}=1-\frac{81}{256}-\frac{9}{16}-\frac{1}{64}=\frac{27}{256} .
\end{array}
$$