11. Given $\frac{x y}{x+y}=2, \frac{x z}{x+z}=3, \frac{y z}{y+z}=4$. Find the value of $7 x+5 y-2 z$.
Three, 11. Given
$$
\begin{array}{l}
\frac{1}{2}=\frac{1}{x}+\frac{1}{y}, \frac{1}{3}=\frac{1}{x}+\frac{1}{z}, \\
\frac{1}{4}=\frac{1}{y}+\frac{1}{z} .
\end{array}
$$

Solving simultaneously, we get $x=\frac{24}{7}, y=\frac{24}{5}, z=24$.
Therefore, $7 x+5 y-2 z=0$.