11. (16 points) Positive real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
\frac{2}{5} \leqslant z \leqslant \min \{x, y\}, \\
x z \geqslant \frac{4}{15}, \\
y z \geqslant \frac{1}{5} .
\end{array}\right.
$$

Find the maximum value of $\frac{1}{x}+\frac{2}{y}+\frac{3}{z}$.
11. From $\frac{2}{5} \leqslant z \leqslant \min \{x, y\}$, we get
$$
\frac{1}{z} \leqslant \frac{5}{2}, \frac{z}{x} \leqslant 1 \text {. }
$$

Also, from $x z \geqslant \frac{4}{15}$, we get $\frac{1}{\sqrt{x}} \leqslant \frac{\sqrt{15}}{2} \cdot \sqrt{z}$.
$$
\begin{array}{l}
\text { Then } \frac{1}{x}+\frac{1}{z}=\frac{2}{x}+\frac{1}{z}-\frac{1}{x} \\
=\frac{2}{\sqrt{x}} \cdot \frac{1}{\sqrt{x}}+\frac{1}{z}\left(1-\frac{z}{x}\right) \\
\leqslant \frac{2}{\sqrt{x}} \cdot \frac{\sqrt{15}}{2} \cdot \sqrt{z}+\frac{5}{2}\left(1-\frac{z}{x}\right) \\
=-\frac{5}{2}\left(\frac{\sqrt{z}}{\sqrt{x}}-\frac{\sqrt{15}}{5}\right)^{2}+4 \leqslant 4 .
\end{array}
$$

Similarly, from $\frac{1}{\sqrt{y}} \leqslant \sqrt{5} \cdot \sqrt{z}, \frac{z}{y} \leqslant 1$, we get
$$
\begin{array}{l}
\frac{1}{y}+\frac{1}{z}=\frac{2}{y}+\frac{1}{z}-\frac{1}{y} \\
=\frac{2}{\sqrt{y}} \cdot \frac{1}{\sqrt{y}}+\frac{1}{z}\left(1-\frac{z}{y}\right) \\
\leqslant \frac{2}{\sqrt{y}} \cdot \sqrt{5} \cdot \sqrt{z}+\frac{5}{2}\left(1-\frac{z}{y}\right) \\
=-\frac{5}{2}\left(\frac{\sqrt{z}}{\sqrt{y}}-\frac{2}{\sqrt{5}}\right)^{2}+\frac{9}{2} \leqslant \frac{9}{2} . \\
\text { Therefore, } \frac{1}{x}+\frac{2}{y}+\frac{3}{z}=\left(\frac{1}{x}+\frac{1}{z}\right)+2\left(\frac{1}{y}+\frac{1}{z}\right) \\
\leqslant 4+2 \times \frac{9}{2}=13,
\end{array}
$$

When $x=\frac{2}{3}, y=\frac{1}{2}, z=\frac{2}{5}$, the equality holds for all the above expressions.
Thus, the maximum value of $\frac{1}{x}+\frac{2}{y}+\frac{3}{z}$ is 13.