11. Given positive numbers $a, b, c, d, e, f$ satisfy
$$
\begin{array}{l}
\frac{b c d e f}{a}=4, \frac{a c d e f}{b}=9, \frac{a b d e f}{c}=16, \\
\frac{a b c e f}{d}=\frac{1}{4}, \frac{a b c d f}{e}=\frac{1}{9}, \frac{a b c d e}{f}=\frac{1}{16} .
\end{array}
$$

Then the value of $(a+c+e)-(b+d+f)$ is
$\qquad$
11. $-2 \frac{7}{12}$.
Since $(a b c d e f)^{4}=1$, and $a, b, c, d, e, f$ are all positive, we have $a b c d e f=1$.
Thus, $\frac{1}{a^{2}}=4 \Rightarrow a=\frac{1}{2}$.
Similarly, $b=\frac{1}{3}, c=\frac{1}{4}, d=2, e=3, f=4$.
Therefore, $(a+c+e)-(b+d+f)=-2 \frac{7}{12}$.