7. Let $A(0, b)$ be a vertex of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$. It is known that the line $y=x-3$ intersects the ellipse at points $P$ and $Q$, and the centroid of $\triangle A P Q$ is the right focus of the ellipse. Then the coordinates of the right focus of the ellipse are $\qquad$
7. $\left(\frac{27-3 \sqrt{5}}{19}, 0\right)$.

As shown in Figure 7, let $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2}, y_{2}\right)$, and the right focus of the ellipse $F(c, 0)$.
Then, by the centroid formula of $\triangle A P Q$, we have
$$
\begin{array}{l}
c=\frac{x_{1}+x_{2}}{3}, \\
0=\frac{y_{1}+y_{2}+b}{3} .
\end{array}
$$

Thus, $x_{1}+x_{2}=3 c, y_{1}+y_{2}=-b$.
Subtracting $\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=1$ and $\frac{x_{2}^{2}}{a^{2}}+\frac{y_{2}^{2}}{b^{2}}=1$, we get
$$
\begin{array}{l}
\frac{\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}\right)}{a^{2}}+\frac{\left(y_{1}-y_{2}\right)\left(y_{1}+y_{2}\right)}{b^{2}}=0 . \\
\text { Also, } \frac{y_{1}-y_{2}}{x_{1}-x_{2}}=1 \text {, then } \\
\frac{x_{1}+x_{2}}{a^{2}}+\frac{y_{1}+y_{2}}{b^{2}}=0 \\
\Rightarrow \frac{3 c}{a^{2}}+\frac{-b}{b^{2}}=0 \\
\Rightarrow a^{2}=3 b c .
\end{array}
$$

From equation (1), the midpoint $M\left(\frac{3 c}{2},-\frac{b}{2}\right)$ of $P Q$.
Since point $M$ lies on the line $P Q$, we have
$$
-\frac{b}{2}=\frac{3 c}{2}-3 \Rightarrow b+3 c=6 \text {. }
$$

From $a^{2}=b^{2}+c^{2}$ and equations (2) and (3), eliminating $a$ and $b$, we get
$$
\begin{array}{l}
(6-3 c)^{2}+c^{2}=3(6-3 c) c \\
\Rightarrow 19 c^{2}-54 c+36=0 .
\end{array}
$$

Solving, we get $c=\frac{27 \pm 3 \sqrt{5}}{19}$.
But from equation (3), $c<2$, so $c=\frac{27-3 \sqrt{5}}{19}$.
Therefore, the coordinates of the right focus are $\left(\frac{27-3 \sqrt{5}}{19}, 0\right)$.