(9) (14 points) For any point $P$ on the ellipse $C: \frac{x^{2}}{3}+\frac{y^{2}}{2}=1$, draw a perpendicular line $P H$ from $P$ to the right directrix of the ellipse $C$ (where $H$ is the foot of the perpendicular), and extend $P H$ to point $Q$ such that $|H Q|=\lambda|P H|$ $\left(\frac{2}{3} \leqslant \lambda \leqslant 1\right)$. As point $P$ moves on the ellipse $C$, the eccentricity range of the trajectory of point $Q$ is $\left[\frac{\sqrt{3}}{3}, 1\right)$. Find the equation of the trajectory of point $Q$.
9 Let $P\left(x_{1}, y_{1}\right), Q(x, y)$. Since the equation of the right directrix is $x=3$, the coordinates of point $H$ are $(3, y)$. Also, $H Q=\lambda P H$, thus $\frac{H P}{P Q}=\frac{-1}{1+\lambda}$. By the section formula, we get 
$$
\left\{\begin{array}{l}x_{1}=\frac{3(1+\lambda)-x}{\lambda} \\ y_{1}=y,\end{array}\right.
$$
Substituting into the ellipse equation, the trajectory of point $Q$ is
$$
\frac{[x-3(1+\lambda)]^{2}}{3 \lambda^{2}}+\frac{y^{2}}{2}=1.
$$

We consider two cases:
(1) The major axis of the ellipse is parallel to the $x$-axis. By the eccentricity
$$
e=\frac{\sqrt{3 \lambda^{2}-2}}{\sqrt{3} \lambda}=\sqrt{1-\frac{2}{3 \lambda^{2}}} \in\left[\frac{\sqrt{3}}{3}, 1\right),
$$

and $\lambda \leqslant 1$, we solve to get $\lambda=1$. Substituting into the equation
$$
\frac{[x-3(1+\lambda)]^{2}}{3 \lambda^{2}}+\frac{y^{2}}{2}=1,
$$

we get the trajectory equation of point $Q$ as $\frac{(x-6)^{2}}{3}+\frac{y^{2}}{2}=1$.
(2) The major axis of the ellipse is parallel to the $y$-axis. In this case, by the eccentricity
$$
e=\frac{\sqrt{2-3 \lambda^{2}}}{\sqrt{2}}=\sqrt{1-\frac{3 \lambda^{2}}{2}} \in\left[\frac{\sqrt{3}}{3}, 1\right),
$$

and $\lambda \geqslant \frac{2}{3}$, we solve to get $\lambda=\frac{2}{3}$. Thus, the trajectory of point $Q$ is $\frac{(x-5)^{2}}{\frac{4}{3}}+\frac{y^{2}}{2}=1$.
In summary, the trajectory equation of point $Q$ is
$$
\frac{(x-6)^{2}}{3}+\frac{y^{2}}{2}=1
$$

or
$$
\frac{(x-5)^{2}}{\frac{4}{3}}+\frac{y^{2}}{2}=1.
$$