14. Given the ellipse $3 x^{2}+y^{2}=6$, there is an inscribed triangle $P A E$, where the coordinates of point $P$ are
$(1, \sqrt{3})$, and the inclination angles of lines $P A$ and $P B$ are complementary. Find the maximum value of the area of triangle $P A B$.
14. Let the slope of $PA$ be $k$, and the slope of $PB$ be $-k$, so the equation of $PA$ is: $y - \sqrt{3} = k(x - 1)$, and the equation of $PB$ is $y - \sqrt{3} = -k(x - 1)$.

By solving the system of equations:
\[
\left\{
\begin{array}{l}
3x^2 + y^2 = 6 \\
y - \sqrt{3} = k(x - 1)
\end{array}
\right.
\]
we get:
\[
(3 + k^2)x^2 - (2k^2 - 2\sqrt{3}k)x + k^2 - 2\sqrt{3}k - 3 = 0
\]
Solving this, we find:
\[
x_A = \frac{k^2 - 2\sqrt{3}k - 3}{3 + k^2}
\]
Similarly, we get:
\[
x_B = \frac{k^2 + 2\sqrt{3}k - 3}{3 + k^2}
\]

Thus, the equation of line $AB$ can be set as: $y = \sqrt{3}x + m$.
By solving the system of equations:
\[
3x^2 + y^2 - 6 \rightarrow 6x^2 + 2\sqrt{3}mx + m^2 - 6 = 0
\]

Therefore, the length of $|AB|$ is:
\[
|AB| = \sqrt{1 + 3} \cdot \sqrt{(x_1 - x_2)^2 - 4x_1 \cdot x_2}
\]
\[
\begin{array}{l}
-2 \sqrt{\left(\frac{\sqrt{3}m}{3}\right)^2 - 4 \times \frac{m^2 - 6}{6}} \\
= \frac{2}{\sqrt{3}} \sqrt{12m^2}
\end{array}
\]

The distance from point $P(1, \sqrt{3})$ to line $AB$ is:
\[
d = \left| \frac{m}{2} \right|
\]

And the equality holds only when $m^2 = 6$, hence:
\[
\left(S_{\triangle PAB}\right)_{\max} = \sqrt{3}
\]