21. (12 points) Let the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$. Given that $a b=2 \sqrt{3}$, and the eccentricity $e=\frac{\sqrt{6}}{3}$.
(1) Find the equation of the ellipse.
(2) There is a line $l$ with slope $k$ passing through the right focus $F$ of the ellipse, and intersecting the ellipse at points $A$ and $B$. Point $P$ lies on the line $x=3$. Is there such a line $l$ that makes $\triangle A B P$ an equilateral triangle? If it exists, find the equation of the line $l$; if not, explain the reason.
21. (1) According to the problem, $c=2, \frac{c}{a}=\frac{\sqrt{6}}{3}$.

Solving, we get $a^{2}=6, b^{2}=2$.
Thus, the equation of the ellipse is $\frac{x^{2}}{6}+\frac{y^{2}}{2}=1$.
(2) Let the line $l: y=k(x-2)$, and combine it with the equation of the ellipse, eliminating $y$ and rearranging, we get
$$
\left(3 k^{2}+1\right) x^{2}-12 k^{2} x+12 k^{2}-6=0 \text {. }
$$

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Then $x_{1}+x_{2}=\frac{12 k^{2}}{3 k^{2}+1}, x_{1} x_{2}=\frac{12 k^{2}-6}{3 k^{2}+1}$.
Thus, $|A B|=\sqrt{1+k^{2}}\left|x_{1}-x_{2}\right|$
$$
\begin{array}{l}
=\sqrt{\left(1+k^{2}\right)\left[\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}\right]} \\
=\frac{2 \sqrt{6}\left(k^{2}+1\right)}{3 k^{2}+1} .
\end{array}
$$

Let the midpoint of side $A B$ be $M\left(x_{0}, y_{0}\right)$.
Then $x_{0}=\frac{6 k^{2}}{3 k^{2}+1}, y_{0}=-\frac{2 k}{3 k^{2}+1}$.
By the problem, the slope of line $M P$ is $-\frac{1}{k}$, and $x_{P}=3$.
$$
\begin{array}{l}
\text { Thus, }|M P|=\sqrt{1+\frac{1}{k^{2}}}\left|x_{0}-x_{P}\right| \\
=\sqrt{\frac{k^{2}+1}{k^{2}}} \cdot \frac{3\left(k^{2}+1\right)}{3 k^{2}+1} .
\end{array}
$$

Notice that, when $\triangle A B P$ is an equilateral triangle, we have $|M P|=\frac{\sqrt{3}}{2}|A B|$
$\Rightarrow \sqrt{\frac{k^{2}+1}{k^{2}}} \cdot \frac{3\left(k^{2}+1\right)}{3 k^{2}+1}=\frac{\sqrt{3}}{2} \cdot \frac{2 \sqrt{6}\left(k^{2}+1\right)}{3 k^{2}+1}$ $\Rightarrow k= \pm 1$.
Thus, the line $l: x-y-2=0$ or $x+y-2=0$.