\(\frac{5\left(3a-2b\right)}{25}=\frac{3\left(2c-5a\right)}{9}=\frac{2\left(5b-3c\right)}{4}=\frac{15a-10b+6c-15a+10b-6c}{25+9+4}=\frac{0}{38}=0\)
\(\Rightarrow3a-2b=0\Leftrightarrow\frac{a}{2}=\frac{b}{3}\)
\(\Leftrightarrow2c-5a=0\Leftrightarrow\frac{a}{2}=\frac{c}{5}\)
\(\Rightarrow\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=\frac{a+b+c}{2+3+5}=\frac{50}{10}=5\)
a=10
b=15
c=25