10. In the Cartesian coordinate system $x O y$, the ellipse $C: \frac{x^{2}}{3}+y^{2}=1$ has its upper vertex at $A$. A line $l$ intersects the ellipse $C$ at points $P$ and $Q$, and does not pass through point $A$, with $\overrightarrow{A P} \cdot \overrightarrow{A Q}=0$.
(1) Does the line $l$ pass through a fixed point? If so, find the coordinates of the fixed point; if not, explain the reason;
(2) Tangents to the ellipse are drawn through points $P$ and $Q$, intersecting at point $B$. Find the range of the area of $\triangle B P Q$.
(1) Let $l: y=k x+m, P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, combining $\left\{\begin{array}{l}y=k x+m, \\ x^{2}+3 y^{2}=3\end{array}\right.$
$$
\begin{array}{l}
\Rightarrow\left(3 k^{2}+1\right) x^{2}+6 k m x+3 m^{2}-3=0,(*) \\
\Delta=36 k^{2} m^{2}-12\left(3 k^{2}+1\right)\left(m^{2}-1\right)=12\left(3 k^{2}-m^{2}+1\right)>0,
\end{array}
$$

Then $\overrightarrow{A P} \cdot \overrightarrow{A Q}=x_{1} x_{1}+\left(y_{1}-1\right)\left(y_{2}-1\right)=x_{1} x_{2}+\left(k x_{1}+m-1\right)\left(k x_{2}+m-1\right)$
$$
\begin{array}{l}
=\left(k^{2}+1\right) x_{1} x_{2}+k(m-1)\left(x_{1}+x_{2}\right)+(m-1)^{2}=0 \\
\Rightarrow 3\left(k^{2}+1\right)\left(m^{2}-1\right)-6 k^{2} m(m-1)+\left(3 k^{2}+1\right)(m-1)^{2}=0 \\
\Rightarrow 3\left(k^{2}+1\right)(m+1)-6 k^{2} m+\left(3 k^{2}+1\right)(m-1)=0 \Rightarrow m=-\frac{1}{2} .
\end{array}
$$

Thus, the line $l$ passes through the fixed point $\left(0,-\frac{1}{2}\right)$.
(2) From (1), we know the line $l: y=k x-\frac{1}{2}$,

Equation $(*)$ becomes $\left(3 k^{2}+1\right) x^{2}-3 k x-\frac{9}{4}=0$.
By $\left\{\begin{array}{l}x_{1} x+3 y_{1} y=3, \\ x_{2} x+3 y_{2} y=3\end{array} \Rightarrow B(6 k,-2)\right.$.
Thus, $|P Q|=\sqrt{k^{2}+1} \cdot \frac{\sqrt{36 k^{2}+9}}{3 k^{2}+1}$,
The distance from $B$ to the line $l$ is $d=\frac{6 k^{2}+\frac{3}{2}}{\sqrt{k^{2}+1}}$, then $S_{\triangle B P Q}=\frac{1}{2}\left(6 k^{2}+\frac{3}{2}\right) \cdot \frac{3 \sqrt{4 k^{2}+1}}{3 k^{2}+1}=\frac{9}{4} \cdot \frac{\left(4 k^{2}+1\right) \sqrt{4 k^{2}+1}}{3 k^{2}+1}$ $\xlongequal{t=\sqrt{4 k^{2}+1} \geqslant 1} \frac{9 t^{3}}{3 t^{2}+1}=\frac{9}{\frac{3}{t}+\frac{1}{t^{3}}} \in\left[\frac{9}{4},+\infty\right)$.