11. Let $a, b, c$ be distinct positive integers. Then the minimum value of $\frac{a b c}{a+b+c}$ is . $\qquad$
11.1.

Assume $a>b>c$. Then $a \geqslant 3, b \geqslant 2, c \geqslant 1$.
Thus, $a b \geqslant 6, b c \geqslant 2, c a \geqslant 3$.
Therefore, $\frac{a+b+c}{a b c}=\frac{1}{b c}+\frac{1}{c a}+\frac{1}{a b}$
$\leqslant \frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$
$\Rightarrow \frac{a b c}{a+b+c} \geqslant 1$.
When $a=3, b=2, c=1$, the equality holds.
Therefore, the minimum value of $\frac{a b c}{a+b+c}$ is 1.