11. Given the parabola $C: y^{2}=2 p x(p>0)$ intersects with the moving line $l: y=k x+b(k, b$ are parameters, and $k \neq 0$, $b \neq 0)$ at points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, and the origin of the Cartesian coordinate system is $O$. Let the slopes of lines $O A$ and $O B$ be $k_{O A}$ and $k_{O B}$, respectively. If $k_{O A} k_{O B}=\sqrt{3}$ always holds, then when $k$ varies, the fixed point through which line $l$ always passes is ( ).
(A) $(-\sqrt{3} p, 0)$
(B) $(-2 \sqrt{3} p, 0)$
(C) $\left(-\frac{\sqrt{3}}{3} p, 0\right)$
(D) $\left(-\frac{2 \sqrt{3}}{3} p, 0\right)$
11. D.

By combining the equation of the line with the equation of the parabola, and eliminating and rearranging, we get
$$
k^{2} x^{2}+(2 k b-2 p) x+b^{2}=0 .
$$

Thus, $x_{1}+x_{2}=\frac{-2 k b+2 p}{k^{2}}, x_{1} x_{2}=\frac{b^{2}}{k^{2}}$.
Since $k_{O A} k_{O B}=\sqrt{3}$, we have
$$
y_{1} y_{2}=\sqrt{3} x_{1} x_{2} \text {. }
$$
$$
\begin{array}{l}
\text { Then } y_{1} y_{2}=\left(k x_{1}+b\right)\left(k x_{2}+b\right) \\
=k^{2} x_{1} x_{2}+k b\left(x_{1}+x_{2}\right)+b^{2} \\
=\frac{2 b p}{k}=\sqrt{3} x_{1} x_{2}=\frac{\sqrt{3} b^{2}}{k^{2}} .
\end{array}
$$

Since $b \neq 0$, we have $b=\frac{2 p k}{\sqrt{3}}$.
Thus, $y=k x+\frac{2 p k}{\sqrt{3}}=k\left(x+\frac{2}{\sqrt{3}} p\right)$.
Therefore, the line $l$ always passes through the fixed point $\left(-\frac{2 \sqrt{3}}{3} p, 0\right)$.