13. As shown in Figure 1, $A^{\prime}(2,1)$, $B^{\prime}(6,3)$, and $C^{\prime}(3,5)$ are three points on the coordinate plane $x O y$. Points $A$, $B$, and $C$ are the midpoints of line segments $C C^{\prime}$, $A A^{\prime}$, and $B B^{\prime}$, respectively. The coordinates of the centroid of $\triangle A B C$ are $\qquad$
13. $\left(\frac{11}{3}, 3\right)$.

Let $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right) 、 C\left(x_{3}, y_{3}\right)$. By the midpoint formula, we have
$$
\begin{array}{l}
2 x_{1}=x_{3}+3,2 x_{2}=x_{1}+2,2 x_{3}=x_{2}+6, \\
2 y_{1}=y_{3}+5,2 y_{2}=y_{1}+1,2 y_{3}=y_{2}+3 .
\end{array}
$$

Solving these equations, we get $x_{1}=\frac{26}{7}, x_{2}=\frac{20}{7}, x_{3}=\frac{31}{7}$,
$$
y_{1}=\frac{27}{7}, y_{2}=\frac{17}{7}, y_{3}=\frac{19}{7} \text {. }
$$

Let the centroid coordinates of $\triangle A B C$ be $\left(x_{0}, y_{0}\right)$, then
$$
x_{0}=\frac{x_{1}+x_{2}+x_{3}}{3}=\frac{11}{3}, y_{0}=\frac{y_{1}+y_{2}+y_{3}}{3}=3 \text {. }
$$

Therefore, the centroid coordinates of $\triangle A B C$ are $\left(\frac{11}{3}, 3\right)$.