11. In the Cartesian coordinate system $x O y$, on the parabola $\Gamma: y^{2}=2 p x(p>0)$, there are three points $A, B, C$ whose orthocenter is the focus $F$. Let the centroid of $\triangle A B C$ be $G$. Find the range of the slope $k$ of the line $F G$.
Let $A\left(2 p a^{2}, 2 p a\right), B\left(2 p b^{2}, 2 p b\right), C\left(2 p c^{2}, 2 p c\right)$, then $\overrightarrow{F A} \cdot \overrightarrow{C B}=0$
$$
\Rightarrow\left(2 p a^{2}-\frac{p}{2}\right)\left(2 p b^{2}-2 p c^{2}\right)+2 p a(2 p b-2 p c)=0 \Rightarrow\left(4 a^{2}-1\right)(b+c)+4 a=0 \text {. }
$$

Similarly, we get $\left(4 b^{2}-1\right)(c+a)+4 b=\left(4 c^{2}-1\right)(a+b)+4 c$
$$
\Rightarrow 4 b^{2}(c+a)-4 c^{2}(a+b)=5(c-b) \Rightarrow a b+b c+c a=-\frac{5}{4} \text {. }
$$

When $a^{2}=\frac{1}{4}$, the x-coordinate of point $A$ is $\frac{p}{2}$, at this time $B C \perp x$ axis, which is a contradiction. Since $b+c=-\frac{4 a}{4 a^{2}-1}$,
$$
b c=-\frac{5}{4}-a(b+c)=-\frac{5}{4}+\frac{4 a^{2}}{4 a^{2}-1}=-\frac{1}{4}+\frac{1}{4 a^{2}-1} .
$$

Then $(b+c)^{2}>4 b c \Rightarrow \frac{16 a^{2}}{\left(4 a^{2}-1\right)^{2}}>-1+\frac{4}{4 a^{2}-1} \Rightarrow\left(4 a^{2}-1\right)^{2}-4\left(4 a^{2}-1\right)+16 a^{2}>0$
$$
\Rightarrow 16 a^{4}-8 a^{2}+5>0 \Rightarrow\left(4 a^{2}-1\right)^{2}+4>0 \Rightarrow a \in \mathbf{R} \text {. }
$$
The centroid of $\triangle A B C$ is $G\left(\frac{2 p}{3}\left(a^{2}+b^{2}+c^{2}\right), \frac{2 p}{3}(a+b+c)\right)$,
and $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}+\frac{5}{2}$, set $u=a+b+c=a-\frac{4 a}{4 a^{2}-1}$,
thus $k_{F G}=\frac{\frac{2 p}{3} u}{\frac{2 p}{3}\left(u^{2}+\frac{5}{2}\right)-\frac{p}{2}}=\frac{4 u}{4 u^{2}+7}$, when $u=0$, $k_{F G}=0$;
when $u \neq 0$, $\frac{4 u}{4 u^{2}+7}=\frac{4}{4 u+\frac{7}{u}} \in\left[-\frac{\sqrt{7}}{7}, 0\right) \cup\left(0, \frac{\sqrt{7}}{7}\right]$.
In summary, the range of the slope $k$ of line $F G$ is $\left[-\frac{\sqrt{7}}{7}, \frac{\sqrt{7}}{7}\right]$.