11. In the Cartesian coordinate system $x O y$, given three points $A\left(0, \frac{4}{3}\right), B(-1,0), C(1,0)$, the distance from point $P$ to line $B C$ is the geometric mean of the distances from the point to lines $A B$ and $A C$.
(1) Find the equation of the locus of point $P$;
(2) If line $L$ passes through the incenter of $\triangle A B C$ (denoted as $D$), and intersects the locus of point $P$ at exactly 3 points, find the range of the slope $k$ of line $L$.
11. Solution: (1) The equations of lines $AB$, $AC$, and $BC$ are $y=\frac{4}{3}(x+1)$, $y=-\frac{4}{3}(x-1)$, and $y=0$, respectively. The distances from point $P(x, y)$ to $AB$, $AC$, and $BC$ are $d_{1}=\frac{1}{5}|4 x-3 y+4|$, $d_{2}=\frac{1}{5}|4 x+3 y-4|$, and $d_{3}=|y|$, respectively.
According to the problem, $d_{1} d_{2}=d_{3}^{2}$, which gives $\left|16 x^{2}-(3 y-4)^{2}\right|=25 y^{3}$, or $16 x^{2}-(3 y-4)^{2}+25 y^{2}=0$, or $16 x^{2}-(3 y-4)^{2}-25 y^{2}=0$. Simplifying, we get the trajectory equation of point $P$ as the circle $S: 2 x^{2}+2 y^{2}+3 y-2=0$ and the hyperbola $T: 8 x^{2}-17 y^{2}+12 y-8=0$.
(2) From the above, the trajectory of point $P$ consists of two parts:
Circle $S: 2 x^{2}+2 y^{2}+3 y-2=0$, (1)
and hyperbola $T: 8 x^{2}-17 y^{2}+12 y-8=0$.
Since points $B(-1,0)$ and $C(1,0)$ satisfy the conditions of the problem, points $B$ and $C$ lie on the trajectory of point $P$, and the common points of the trajectory curves $S$ and $T$ are only points $B$ and $C$. The incenter $D$ of $\triangle ABC$ also satisfies the conditions of the problem, and from $d_{1}=d_{2}=d_{3}$, we get $D\left(0, \frac{1}{2}\right)$, which lies on circle $S$. Line $L$ passes through $D$ and intersects the trajectory of point $P$ at 3 points, so the slope of $L$ exists. Let the equation of $L$ be $y=k x+\frac{1}{2}$.
(i) When $k=0$, $L$ is tangent to circle $S$ and has a unique common point $D$; at this time, the line $y=\frac{1}{2}$ is parallel to the $x$-axis, indicating that $L$ intersects the hyperbola at two points different from $D$, so $L$ intersects the trajectory of point $P$ at exactly 3 points.
(ii) When $k \neq 0$, $L$ intersects circle $S$ at two different points. In this case, $L$ intersects the trajectory of point $P$ at exactly 3 points in only two scenarios:

Scenario 1: Line $L$ passes through point $B$ or point $C$, at which time the slope of $L$ is $k= \pm \frac{1}{2}$, and the equation of line $L$ is $x= \pm(2 y-1)$. Substituting into equation (2) gives $y(3 y-4)=0$, solving for $E\left(\frac{5}{3}, \frac{4}{3}\right)$ or $E\left(-\frac{5}{3}, \frac{4}{3}\right)$. This indicates that line $BD$ intersects hyperbola $T$ at 2 points $B$ and $E$; line $CD$ intersects hyperbola $T$ at 2 points $C$ and $F$.
Thus, when $k= \pm \frac{1}{2}$, $L$ intersects the trajectory of point $P$ at exactly 3 points.
Scenario 2: Line $L$ does not pass through points $B$ and $C$ (i.e., $k \neq \pm \frac{1}{2}$), because $L$ intersects $S$ at two different points, so $L$ intersects hyperbola $T$ at exactly one point. That is, the system of equations $\left\{\begin{array}{l}8 x^{2}-17 y^{2}+12 y-8=0 \\ y=k x+\frac{1}{2}\end{array}\right.$ has exactly one real solution. Eliminating $y$ and simplifying gives $\left(8-17 k^{2}\right) x^{2}-5 k x-\frac{25}{4}=0$. The necessary and sufficient condition for this equation to have a unique real solution is $8-17 k^{2}=0$, (4) or $(-5 k)^{2}+4\left(8-17 k^{2}\right) \frac{25}{4}=0$,
Solving equation (4) gives $k= \pm \frac{2 \sqrt{34}}{17}$, solving equation (5) gives $k= \pm \frac{\sqrt{2}}{2}$. In summary, the set of possible values for the slope $k$ of line $L$ is the finite set $\left\{0, \pm \frac{1}{2}, \pm \frac{2 \sqrt{34}}{17}, \pm \frac{\sqrt{2}}{2}\right\}$.