7. In triangle $A B C$, the altitudes $A D, B E, C F$ are drawn. The length of side $B C$ is 6. The distances from the center of the inscribed circle of triangle $D E F$ to points $B$ and $C$ are $3 \sqrt{2}-\sqrt{6}$ and $2 \sqrt{6}$, respectively. Find the height of triangle $D E F$ drawn to side $D E$. (16 points)
Solution. $\quad A D, B E, C F$ are the altitudes of triangle $A B C, D A, E B, F C$ are the angle bisectors of angles $D, E, F$ of triangle $D E F, O$ is the point of intersection of the altitudes of triangle $A B C$, which is also the center of the inscribed circle of triangle $D E F$. Thus, $B O=3 \sqrt{2}-\sqrt{6}, C O=2 \sqrt{6}$. Let $O D=x, B D=y$. Then we arrive at the system $\left\{\begin{array}{c}x^{2}+y^{2}=(3 \sqrt{2}-\sqrt{6})^{2}, \\ (6-y)^{2}+x^{2}=24 .\end{array}\right.$ Solving the system, we get $y=3-\sqrt{3}, x=3-\sqrt{3}$. Then $\angle E B C=\angle B C A=45^{\circ}$, $E C=3 \sqrt{2}, \quad \angle F C A=\arccos \frac{O E}{O C}=\frac{3 \sqrt{2}}{2 \sqrt{6}}=30^{\circ}$,

![](https://cdn.mathpix.com/cropped/2024_05_06_d66465e084edd0e7e96fg-17.jpg?height=417&width=602&top_left_y=1682&top_left_x=1138)
$\angle A B E=30^{\circ}, \angle D F E=90^{\circ}, F E=B C \cos 60^{\circ}=3$. $\angle F E D=30^{\circ}, h=F E \sin 30^{\circ}=1.5$.

Answer: 1.5 .