Ta có: \(2a=3b\Rightarrow\frac{a}{3}=\frac{b}{2}\left(1\right)\)
\(5b=7c\Rightarrow\frac{b}{7}=\frac{c}{5}\left(2\right)\)
Từ 1 và 2 \(\Rightarrow\frac{a}{21}=\frac{b}{14}=\frac{c}{10}\)
\(\Rightarrow\frac{3a}{63}=\frac{7b}{98}=\frac{5c}{50}=\frac{3a-7b+5c}{63-98+50}=\frac{-30}{15}=-2\)
\(\Rightarrow\frac{a}{21}=-2\Rightarrow a=-42\)
\(\frac{b}{14}=-2\Rightarrow b=-28\)
\(\frac{c}{10}=-2\Rightarrow c=-20\)
Vậy \(a+b+c=-42-28-20=-90\)