Three, (25 points) Let $a$, $b$, $c$ be distinct integers from 1 to 9. Find the maximum possible value of $\frac{a+b+c}{a b c}$.

保留源文本的换行和格式，翻译结果如下：

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Three, (25 points) Let $a$, $b$, $c$ be distinct integers from 1 to 9. Find the maximum possible value of $\frac{a+b+c}{a b c}$.
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Three, let $P=\frac{a+b+c}{a b c}$.
In equation (1), let $a$ and $b$ remain unchanged, and only let $c$ vary, where $c$ can take any integer from 1 to 9.
Then, from $P=\frac{a+b+c}{a b c}=\frac{1}{a b}+\frac{a+b}{a b c}$, we know that
when $c=1$, $P$ reaches its maximum value, so $c=1$.
Thus, $P=\frac{a+b+1}{a b}=\frac{1}{a}+\frac{a+1}{a b}$.
In equation (2), let $a$ remain unchanged, and only let $b$ vary, where $b$ can take any integer from 2 to 9.

Then, from equation (2), we know that when $b=2$, $P$ reaches its maximum value, so $b=2$.
At this point, $P=\frac{1}{2}+\frac{3}{2 a}$.
In equation (3), $P$ reaches its maximum value when $a$ takes its minimum value, and $a$ can take any integer from 3 to 9, so $a=3$.

Therefore, when $a=3, b=2, c=1$ (the letters can be interchanged), $P$ reaches its maximum value of 1.
(Shen Xueming, Changqiao Middle School, Wuzhong District, Suzhou City, Jiangsu Province, 215128)