3. Let $G$ be the centroid of $\triangle ABC$. Determine the position of point $P$ in the plane of $\triangle ABC$ such that $AP \cdot AG + BP \cdot BG + CP \cdot CG$ is minimized, and express this minimum value in terms of the side lengths of $\triangle ABC$.

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Let $a, b, c$ be the lengths of the sides opposite to vertices $A, B, C$ of $\triangle ABC$, respectively. We will prove that
$$
A P \cdot A G + B P \cdot B G + C P \cdot C G
$$

attains its minimum value when $P$ is the centroid $G$, and the minimum value is
$$
\begin{array}{l}
A G^{2} + B G^{2} + C G^{2} \\
=\frac{1}{9}\left[\left(2 b^{2} + 2 c^{2} - a^{2}\right) + \left(2 c^{2} + 2 a^{2} - b^{2}\right) + \left(2 c^{2} + 2 a^{2} - b^{2}\right)\right] \\
=\frac{1}{3}\left(a^{2} + b^{2} + c^{2}\right).
\end{array}
$$

Let $S$ be the circle passing through $B, G, C$, and the median $AL$ intersects $S$ at $G$ and $K$, as shown in Figure 6. Let $\angle B G K = \theta, \angle A G N = \varphi, \angle B G N = \delta$, where $N$ is the midpoint of $AB$. Since $L$ is the midpoint of $BC$, the distances from $B$ and $C$ to $GL$ are equal, i.e., $B G \sin \theta = C G \sin \varphi$. Thus, we have $\frac{B G}{\sin \varphi} = \frac{C G}{\sin \theta}$.
Similarly, we get $\frac{A G}{\sin \delta} = \frac{B G}{\sin \varphi}$.
Since $B K = 2 R \sin \theta, C K = 2 R \cdot \sin \varphi, B C = 2 R \cdot \sin \delta$, where $R$ is the radius of circle $S$, we have
$$
\frac{A \dot{G}}{B C} = \frac{B G}{C K} = \frac{C G}{B K}.
$$

Let $P$ be any point in the plane of $\triangle ABC$. By the generalized Ptolemy's theorem, we have $P K \cdot B C \leqslant B P \cdot C K + C P \cdot B K$, with equality if and only if $P$ lies on circle $S$.

From equation (1), we have $P K \cdot A G \leqslant B P \cdot B G + C P \cdot C G$. Adding $A P \cdot A G$ to both sides, we get
$(A P + P K) \cdot A G \leqslant A P \cdot A G + B P \cdot B G + C P \cdot C G$.
Since $A K \leqslant A P + P K$, we have
$A K \cdot A G \leqslant A P \cdot A G + B P \cdot B G + C P \cdot C G$.
Equality holds if and only if $P$ lies on segment $A K$ and $P$ lies on circle $S$, i.e., equality holds if and only if $P = G$.