2. 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 $|H Q|=\lambda|P H|(\lambda \geqslant 1)$. As point $P$ moves along the ellipse $C$, the range of the eccentricity of the locus of point $Q$ is $\qquad$.
2. $\left[\frac{\sqrt{3}}{3}, 1\right)$ Detailed Explanation: 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, since $H Q=\lambda P H$, we have $\frac{H P}{P Q}=\frac{-\lambda}{1+\lambda}$, so 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 of point $Q$ as $\frac{[x-3(1+\lambda)]^{2}}{3 \lambda^{2}}+\frac{y^{2}}{2}=1$, so the eccentricity $e=\frac{\sqrt{3 \lambda^{2}-2}}{\sqrt{3} \lambda}=\sqrt{1-\frac{2}{3 \lambda^{2}}} \in\left[\frac{\sqrt{3}}{3}, 1\right)$.