7. For any point $P$ on the ellipse $C: \frac{x^{2}}{3}+\frac{y^{2}}{2}=1$, draw a perpendicular line $P H$ (where $H$ is the foot of the perpendicular) to the directrix of $C$. Extend $P H$ to $Q$ such that $H Q=\lambda P H(\lambda \geqslant 1)$. As point $P$ moves along $C$, the range of the eccentricity of the locus of point $Q$ is $\qquad$.
7. $\left[\frac{\sqrt{3}}{3} \cdot 1\right)$

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 $\left(3, y_{1}\right)$.
Also, since $H Q=\lambda P H$, we have $\frac{H P}{P Q}=-\frac{1}{1+\lambda}$. Therefore, 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, we get the trajectory equation of point $Q$ as: $\frac{(x-3(1+\lambda))^{2}}{3 \lambda^{2}}+\frac{y^{2}}{2}=1$, so the eccentricity $e=\frac{\sqrt{3 \lambda^{2}-2}}{\sqrt{3} \lambda}=\sqrt{1-\frac{2}{3 \lambda^{2}}} \in\left[\frac{\sqrt{3}}{3}, 1\right)$.