6. 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 on the ellipse $C$, the range of the eccentricity of the trajectory of point $Q$ is ( ).
(A) $\left(0, \frac{\sqrt{3}}{3}\right]$
(B) $\left(\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{2}\right]$
(C) $\left[\frac{\sqrt{3}}{3}, 1\right)$
(D) $\left(\frac{\sqrt{3}}{2}, 1\right)$
6.C.

Let $P\left(x_{1}, y_{1}\right), Q(x, y)$.
Since the equation of the right directrix is $x=3$, we have $H(3, y)$.
Also, $H Q=\lambda P H$, then
$$
\frac{H P}{P Q}=\frac{-1}{1+\lambda} \text {. }
$$

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 equation of point $Q$ is
$$
\frac{[x-3(1+\lambda)]^{2}}{3 \lambda^{2}}+\frac{y^{2}}{2}=1 \text {. }
$$

Therefore, the eccentricity $e=\frac{\sqrt{3 \lambda^{2}-2}}{\sqrt{3} \lambda}=\sqrt{1-\frac{2}{3 \lambda^{2}}}$
$$
\in\left[\frac{\sqrt{3}}{3}, 1\right) \text {. }
$$