(3) Let $P$ be any point inside $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively, and $S$ is the area of $\triangle ABC$. Then the minimum value of $\frac{a \cdot PA + b \cdot PB + c \cdot PC}{S}$ is $\qquad$
(3) 4 Hint: Since $P A+h_{1} \geqslant h$, then
$$
\begin{array}{c}
a\left(P A+h_{1}\right) \geqslant a h=2 S, \\
a P A+a h_{1} \geqslant 2 S,
\end{array}
$$

Similarly, $b P B+b h_{2} \geqslant 2 S, c P C+c h_{3} \geqslant 2 S$,
Since $a h_{1}+b h_{2}+c h_{3}=2 S$, then
$$
a P A+b P B+c P C \geqslant 4 S,
$$

Equality holds if and only if $P$ is the orthocenter of $\triangle A B C$, hence the minimum value sought is 4.