5. Given $\frac{x-a-b}{c}+\frac{x-b-c}{a}+\frac{x-c-a}{b}$ $=3$, and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \neq 0$. Then $x-a-b-c=$
5. 0 ;

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