jmc

From the given equations, $$\begin{array}{rl}a-b& =\frac{1}{c}-\frac{1}{b}=\frac{b-c}{bc},\\ b-c& =\frac{1}{a}-\frac{1}{c}=\frac{c-a}{ac},\\ c-a& =\frac{1}{b}-\frac{1}{a}=\frac{a-b}{ab}.\end{array}$$Multiplying these equations, we get $$(a-b)(b-c)(c-a)=\frac{(a-b)(b-c)(c-a)}{a^2{b}^2{c}^2}.$$Since $a,$ $b,$ and $c$ are distinct, we can cancel the factors of $a-b,$ $b-c,$ $c-a,$ to get $$a^2{b}^2{c}^2=1.$$Therefore, $|abc|=\boxed{1}.$