10. (This sub-question is worth 15 points) Given an ellipse with its center at the origin $O$, foci on the $x$-axis, eccentricity $\frac{\sqrt{3}}{2}$, and passing through the point $\left(\sqrt{2}, \frac{\sqrt{2}}{2}\right)$. Let a line $l$ that does not pass through the origin $O$ intersect the ellipse at points $P$ and $Q$, and the slopes of the lines $O P, P Q, O Q$ form a geometric sequence. Find the range of the area of $\triangle O P Q$.
Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, then $\left\{\begin{array}{l}\frac{c}{a}=\frac{\sqrt{3}}{2}, \\ \frac{2}{a^{2}}+\frac{1}{2 b^{2}}=1\end{array}\right.$, solving this gives $\left\{\begin{array}{l}a=2 \\ b=1\end{array}\right.$, hence the equation of the ellipse is
$$
\frac{x^{2}}{4}+y^{2}=1
$$

From the problem, we know that the slope of the line $l$ exists and is not 0, so we can set the equation of the line $l$ as $y=k x+m(m \neq 0)$. Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, then $x_{1} \neq x_{2}$ and $x_{1} x_{2} \neq 0$. From $\left\{\begin{array}{l}y=k x+m \\ x^{2}+4 y^{2}-4=0\end{array} \quad\right.$, eliminating $y$ gives
$$
\left(1+4 k^{2}\right) x^{2}+8 k m x+4\left(m^{2}-1\right)=0
$$

Then we have
$$
\Delta=64 k^{2} m^{2}-16\left(1+4 k^{2}\right)\left(m^{2}-1\right)=16\left(4 k^{2}-m^{2}+1\right)>0
$$

And
$$
x_{1}+x_{2}=\frac{-8 k m}{1+4 k^{2}}, x_{1} x_{2}=\frac{4\left(m^{2}-1\right)}{1+4 k^{2}}, y_{1} y_{2}=\left(k x_{1}+m\right)\left(k x_{2}+m\right)=k^{2} x_{1} x_{2}+k m\left(x_{1}+x_{2}\right)+m^{2}
$$

From the slopes of the lines $O P, P Q, O Q$ forming a geometric sequence, we get
$$
\frac{y_{1}}{x_{1}} \cdot \frac{y_{2}}{x_{2}}=\frac{k^{2} x_{1} x_{2}+k m\left(x_{1}+x_{2}\right)+m^{2}}{x_{1} x_{2}}=k^{2}
$$

Thus
$$
\frac{-8 k^{2} m^{2}}{1+4 k^{2}}+m^{2}=0
$$

By $m \neq 0$, we get $k^{2}=\frac{1}{4}, k= \pm \frac{1}{2}$.
10 points
From $\Delta>0$ and $x_{1} x_{2} \neq 0$, we know $0<m^{2}<2$ and $m^{2} \neq 1$.
Let $d$ be the distance from point $O$ to the line $l$, then
$$
S_{\triangle O P Q}=\frac{1}{2} d|P Q|=\frac{1}{2} \frac{|m|}{\sqrt{1+k^{2}}} \sqrt{1+k^{2}}\left|x_{1}-x_{2}\right|=\frac{1}{2}|m| \sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}=\sqrt{m^{2}\left(2-m^{2}\right)}
$$

Therefore, the range of $S_{\triangle O P Q}$ is $(0,1)$.
15 points