11. In the Cartesian coordinate system, the hyperbola $\Gamma: \frac{x^{2}}{3}-y^{2}=1$. For any point $P$ in the plane that is not on $\Gamma$, let $\Omega_{P}$ be the set of all lines passing through point $P$ and intersecting $\Gamma$ at two points. For any line $l \in \Omega_{P}$, let $M, N$ be the two intersection points of $l$ and $\Gamma$, and define $f_{P}(l)=|P M| \cdot|P N|$. If there exists a line $l_{0} \in \Omega_{P}$ such that: the two intersection points of $l_{0}$ and $\Gamma$ are on opposite sides of the $y$-axis, and for any line $l \in \Omega_{P}, l \neq l_{0}$, we have $f_{P}(l)>f_{P}\left(l_{0}\right)$, then $P$ is called a "good point". Find the area of the region formed by all good points.
Let $P(a, b)$ be a good point, and let the line $l \in \Omega_{P}:\left\{\begin{array}{l}x=a+t \cos \theta, \\ y=b+t \sin \theta\end{array}\right.$ (where $t$ is a parameter, and $\theta$ is the inclination angle of the line $l$). Substituting into the hyperbola $\Gamma: x^{2}-3 y^{2}=3$, we get $(t \cos \theta+a)^{2}-3(t \sin \theta+b)^{2}-3=0$ $\Rightarrow\left(\cos ^{2} \theta-3 \sin ^{2} \theta\right) t^{2}+(2 a \cos \theta-6 b \sin \theta) t+a^{2}-3 b^{2}-3=0$.
The above equation has two distinct real solutions $t_{1}, t_{2}$, then $\left\{\begin{array}{l}\cos ^{2} \theta-3 \sin ^{2} \theta \neq 0, \\ \Delta>0\end{array}\right.$
$\Rightarrow\left\{\begin{array}{l}\theta \neq \frac{\pi}{6}, \frac{5 \pi}{6} \\ (b \cos \theta-a \sin \theta)^{2}>3 \sin ^{2} \theta-\cos ^{2} \theta .\end{array}\right.$
When the inclination angle $\theta$ of $l$ satisfies (1), it is known that $f_{P}(l)=\left|t_{1} t_{2}\right|=\frac{\left|a^{2}-3 b^{2}-3\right|}{\left|\cos ^{2} \theta-3 \sin ^{2} \theta\right|}$.
Clearly, $P$ is not on $\Gamma$, so $a^{2}-3 b^{2}-3 \neq 0$. When $l$ intersects $\Gamma$ at two points on opposite sides of the $y$-axis, $\theta \in \left[0, \frac{\pi}{6}\right) \cup\left(\frac{5 \pi}{6}, \pi\right)$, at this time $3 \sin ^{2} \theta-\cos ^{2} \theta\left|a^{2}-3 b^{2}-3\right| \Rightarrow 0<3 \sin ^{2} \theta-\cos ^{2} \theta<1$ $\Rightarrow \theta \in\left(\frac{\pi}{6}, \frac{\pi}{4}\right) \cup\left(\frac{3 \pi}{4}, \frac{5 \pi}{6}\right)$. Therefore, for any $\theta \in\left[\frac{\pi}{4}, \frac{3 \pi}{4}\right]$, (1) does not hold, i.e., for any $\theta \in\left[\frac{\pi}{4}, \frac{3 \pi}{4}\right]$, we have $|b \cos \theta-a \sin \theta| \leqslant \sqrt{3 \sin ^{2} \theta-\cos ^{2} \theta}$.
In (2), let $\theta=\frac{\pi}{4}, \frac{3 \pi}{4}$, we get $\left|\frac{\sqrt{2}}{2} b-\frac{\sqrt{2}}{2} a\right| \leqslant 1,\left|-\frac{\sqrt{2}}{2} b-\frac{\sqrt{2}}{2} a\right| \leqslant 1$
$\Rightarrow\left\{\begin{array}{l}|a-b| \leqslant \sqrt{2} \\ |a+b| \leqslant \sqrt{2}\end{array} \Rightarrow|a|+|b| \leqslant \sqrt{2}\right.$.
Conversely, when $|a|+|b| \leqslant \sqrt{2}$, $\theta \in\left[\frac{\pi}{4}, \frac{3 \pi}{4}\right] \Rightarrow \sin \theta \geqslant|\cos \theta|$
$\Rightarrow|b \cos \theta-a \sin \theta| \leqslant|b||\cos \theta|+|a| \sin \theta \leqslant(|a|+|b|) \sin \theta \leqslant \sqrt{2} \sin \theta$,
and $\sqrt{3 \sin ^{2} \theta-\cos ^{2} \theta} \geqslant \sqrt{3 \sin ^{2} \theta-\sin ^{2} \theta}=\sqrt{2} \sin \theta$ shows that (2) holds.
Therefore, $P(a, b)$ is a good point if and only if $|a|+|b| \leqslant \sqrt{2}$, thus all good points correspond to the region $A=\{(a, b)|| a|+| b \mid \leqslant \sqrt{2}\}$.
In summary, the area $S_{A}=4$ of the region formed by all good points.