3. From a point $P$ inside $\triangle A B C$, perpendiculars are drawn to the sides $B C, C A, A B$, with the feet of the perpendiculars being $D, E, F$ respectively. Semi-circles are constructed outwardly on $A F, B F$, $B D, C D, C E, A E$ as diameters. As shown in the figure, the areas of these six semi-circles are denoted as $S_{1}, S_{2}$, $S_{3}, S_{4}, S_{5}, S_{6}$, respectively. If $S_{5}-S_{6}=2, S_{1}-S_{2}=1$, then $S_{4}-S_{3}=$ $\qquad$
Solution: As shown in the figure,
, connect $A P, B P, C P$. According to the Pythagorean theorem, we can get $A P^{2}-B P^{2}=A F^{2}-B F^{2}, B P^{2}-C P^{2}=B D^{2}-C D^{2}, C P^{2}-A P^{2}=C E^{2}-E A^{2}$, so $A F^{2}+B D^{2}+C E^{2}=B F^{2}+C D^{2}+E A^{2}$; then according to the formula for the area of a circle, we can get $S_{1}+S_{3}+S_{5}=S_{2}+S_{4}+S_{6}$, so $S_{4}-S_{3}=\left(S_{5}-S_{6}\right)+\left(S_{1}-S_{2}\right)=2+1=3$, solve accordingly.

[Answer] Solution: As shown in the figure,
, connect $A P, B P, C P$,
According to the Pythagorean theorem, we can get $A P^{2}=A F^{2}+F P^{2} \cdots$ (1), $B P^{2}=B F^{2}+F P^{2} \cdots$ (2),
(1) - (2), we can get $A P^{2}-B P^{2}=A F^{2}-B F^{2} \cdots$ (3),
Similarly, we can get $B P^{2}-C P^{2}=B D^{2}-C D^{2} \cdots$ (4),
Similarly, we can get $C P^{2}-A P^{2}=C E^{2}-E A^{2} \cdots$ (5),
(3)+(4)+(5), we can get $A F^{2}+B D^{2}+C E^{2}=B F^{2}+C D^{2}+E A^{2}$,
So $\frac{\pi}{4} \times\left(A F^{2}+B D^{2}+C E^{2}\right)=\frac{\pi}{4} \times\left(B F^{2}+C D^{2}+E A^{2}\right)$,
Therefore, $S_{1}+S_{3}+S_{5}=S_{2}+S_{4}+S_{6}$,
So $S_{4}-S_{3}$
$$
\begin{array}{l}
=\left(S_{5}-S_{6}\right)+\left(S_{1}-S_{2}\right) \\
=2+1 \\
=3
\end{array}
$$

The answer is: 3.