11. $\sqrt{x^{2}+y^{2}-2 x-2 y+2}+\sqrt{x^{2}+y^{2}-2 x+4 y+2 \sqrt{3} y+8+4 \sqrt{3}}+$ $\sqrt{x^{2}+y^{2}+8 x+4 \sqrt{3} x-4 y+32+16 \sqrt{3}}(x, y \in \mathbf{R})$ The minimum value is $\qquad$
$$
\begin{array}{l}
\text { 11. } 3 \sqrt{6}+4 \sqrt{2} \\
\text { Let } A(1,1), B(1,-2-\sqrt{3}), C(-4-2 \sqrt{3} \cdot 2), P(x, y), \text { then } f(x, y)= \\
|P A|+|P B|+|P C| \text {. }
\end{array}
$$
11. $3 \sqrt{6}+4 \sqrt{2}$

Let $A(1,1), B(1,-2-\sqrt{3}), C(-4-2 \sqrt{3} \cdot 2), P(x, y)$, then $f(x, y)=$ $|P A|+|P B|+|P C|$.
Notice that $\angle A O B=\angle A O C=\angle B O C=120^{\circ}$, so $O$ is the Fermat point of $\triangle A B C$.
Therefore, the minimum value of $f$ is obtained when $P=O$, i.e., $|O A|+|O B|+|O C|=\sqrt{2}+\sqrt{6}+\sqrt{2}+$
$$
2 \sqrt{6}+2 \sqrt{2}=3 \sqrt{6}+4 \sqrt{2} \text {. }
$$