If \(\frac{6 a-4}{a}+\frac{6 b-4}{b}+\frac{6 c-4}{c}=0\), then \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=?\)

1
\(\frac{9}{2}\)
2
\(\frac{2}{9}\)
3
\(\frac{4}{3}\)
4
\(\frac{5}{8}\)

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