3. $P$ is a moving point on the plane of the equilateral $\triangle A B C$ with side length 2, and $P A^{2}+P B^{2}+P C^{2}=16$. Then the trajectory of the moving point $P$ is $\qquad$ .
3. A circle with the center of $\triangle A B C$ and a radius of 2.
Let $O$ be the center of the equilateral $\triangle A B C$, $\angle P O C=\alpha$, then
$$
\angle A O C=\angle B O C=
$$
$120^{\circ}$.
$$
\begin{array}{l}
\quad O A=O B=O C=R= \\
\frac{2}{3} \sqrt{3} .
\end{array}
$$

In $\triangle O A P, \triangle O B P$, and
$\triangle O C P$, by the cosine rule we get
$$
\begin{array}{l}
P A^{2}=O A^{2}+O P^{2}-2 O A \cdot O P \cos \left(120^{\circ}-\alpha\right), \\
P B^{2}=O B^{2}+O P^{2}-2 O B \cdot O P \cos \left(120^{\circ}+\alpha\right), \\
P C^{2}=O C^{2}+O P^{2}-2 O C \cdot O P \cos \alpha .
\end{array}
$$

By the sum-to-product formulas, it is easy to get
$$
\begin{array}{l}
\cos \left(120^{\circ}-\alpha\right)+\cos \left(120^{\circ}+\alpha\right)+\cos \alpha=0 . \\
\therefore P A^{2}+P B^{2}+P C^{2} \\
\quad=3 O P^{2}+O A^{2}+O B^{2}+O C^{2} . \\
\therefore O P=2 .
\end{array}
$$

Therefore, the locus of point $P$ is: a circle with the center of the equilateral $\triangle A B C$ and a radius of 2.