Three. (20 points) 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 $l$ that does not pass through point $A$ intersects the ellipse $C$ at points $P$ and $Q$, and $\overrightarrow{A P} \cdot \overrightarrow{A Q}=0$.
(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) Since $\overrightarrow{A P} \cdot \overrightarrow{A Q}=0$, we have $\overrightarrow{A P} \perp \overrightarrow{A Q}$. When the lines $A P$ and $A Q$ are parallel to the $x$-axis, point $P$ or $Q$ coincides with $A$, which is not allowed.
Let the line $l_{P A}: y=k x+1$. Then
$l_{Q A}: y=-\frac{1}{k} x+1$.
Substituting $y=k x+1$ into $x^{2}+3 y^{2}=3$, we get
$\left(1+3 k^{2}\right) x^{2}+6 k x=0$
$\Rightarrow x_{P}=-\frac{6 k}{1+3 k^{2}}, y_{P}=\frac{2}{1+3 k^{2}}-1$.
Similarly, $x_{Q}=\frac{6 k}{k^{2}+3}, y_{Q}=1-\frac{6}{k^{2}+3}$.
Substituting into the line $l$ gives $\frac{y-y_{P}}{y_{Q}-y_{P}}=\frac{x-x_{P}}{x_{Q}-x_{P}}$.
Simplifying, we get $l: y=\frac{k^{2}-1}{4 k} x-\frac{1}{2}$.
Thus, the line $l$ passes through the fixed point $\left(0,-\frac{1}{2}\right)$.
(2) From (1), we have $A P=\frac{6|k| \sqrt{1+k^{2}}}{1+3 k^{2}}$.
Similarly, $A Q=\frac{6 \sqrt{1+k^{2}}}{k^{2}+3}$.
Therefore, $P Q^{2}=36\left(1+k^{2}\right)\left(\frac{k^{2}}{\left(1+3 k^{2}\right)^{2}}+\frac{1}{\left(k^{2}+3\right)^{2}}\right)$ $=\frac{36\left(1+k^{2}\right)\left(k^{6}+15 k^{4}+15 k^{2}+1\right)}{\left(3 k^{4}+10 k^{2}+3\right)^{2}}$.
Assume $k>0$, let $t=k+\frac{1}{k}$, then $t \geqslant 2$.
Thus, $P Q^{2}=\frac{36 t^{2}\left(t^{2}+12\right)}{\left(3 t^{2}+4\right)^{2}}$, i.e.,
$$
P Q=\frac{6 t \sqrt{t^{2}+12}}{3 t^{2}+4} \text {. }
$$

Let $B\left(x_{0}, y_{0}\right)$. Then the chord of tangents $P Q$ is the equation of $l$: $x_{0} x+3 y_{0} y=3$.
Comparing with (1), we get
$$
y_{0}=-2, x_{0}=\frac{3\left(k^{2}-1\right)}{2 k} \text {. }
$$

Thus, the distance from point $B$ to $P Q$ is
$$
d=\frac{3\left(\frac{k^{2}-1}{k}\right)^{2}+12}{2 \sqrt{\left(\frac{k^{2}-1}{k}\right)^{2}+16}}=\frac{3 t^{2}}{2 \sqrt{t^{2}+12}} \text {. }
$$

Therefore, the area of $\triangle B P Q$ is
$$
S_{\triangle B P Q}=\frac{1}{2} d \times P Q=\frac{9 t^{3}}{2\left(3 t^{2}+4\right)}
$$
$$
\begin{array}{l}
=\frac{9}{2\left(3 t^{-1}+4 t^{-3}\right)} \\
\geqslant \frac{9}{2\left(3 \times 2^{-1}+4 \times 2^{-3}\right)}=\frac{9}{4},
\end{array}
$$

with equality holding if and only if $t=2$, i.e., $k=1$.
Thus, the range of the area of $\triangle B P Q$ is
$$
\left[\frac{9}{4},+\infty\right)
$$