Given $\triangle A B C$ with sides $B C=a, C A=$ $b, A B=c$, and point $Q$ inside the triangle. Let
$$
f=a Q A^{2}+b Q B^{2}+c Q C^{2} \text {. }
$$
(1) Find the minimum value of $f$;
(2) Determine the geometric position of point $Q$ when $f$ takes its minimum value.
(1) As shown in Figure 3, construct the symmetric points of point $Q$ with respect to $BC$, $CA$, and $AB$, denoted as $Q_{1}$, $Q_{2}$, and $Q_{3}$.
Then $S_{\triangle Q_{1}} + S_{\triangle Q_{Q}}$
$$
\begin{array}{l}
=\frac{1}{2} Q A^{2} \left(\sin 2 \alpha_{1} + \sin 2 \beta_{1}\right) \\
= Q A^{2} \sin \left(\alpha_{1} + \beta_{1}\right) \cdot \cos \left(\alpha_{1} - \beta_{1}\right)
\end{array}
$$
$\leqslant Q A^{2} \sin A$.
Similarly, $S_{\triangle B Q Q_{1}} + S_{\triangle B Q Q_{3}} \leqslant Q B^{2} \sin B$,
$S_{\triangle C Q Q_{1}} + S_{\triangle C Q Q_{2}} \leqslant Q C^{2} \sin C$.
Adding these three inequalities, we get
$2 S_{\triangle M B C} \leqslant \frac{a}{2 R} Q A^{2} + \frac{b}{2 R} Q B^{2} + \frac{c}{2 R} Q C^{2}$.
Thus, $f = a Q A^{2} + b Q B^{2} + c Q C^{2}$
$\geqslant 4 R \cdot S_{\triangle A B C} = a b c$.
(2) When the above inequality holds with equality, i.e., when $f$ attains its minimum value, we have
$$
\begin{array}{l}
\cos \left(\alpha_{1} - \beta_{1}\right) = \cos \left(\alpha_{2} - \beta_{2}\right) \\
= \cos \left(\alpha_{3} - \beta_{3}\right) \\
= 1 .
\end{array}
$$

At this point, $\alpha_{1} = \beta_{1}, \alpha_{2} = \beta_{2}, \alpha_{3} = \beta_{3}$, meaning $Q A$, $Q B$, and $Q C$ bisect $\angle A$, $\angle B$, and $\angle C$, respectively. Therefore, $Q$ must be the incenter of $\triangle A B C$.