\(\dfrac{a+b}{3}=\dfrac{b+c}{4}=\dfrac{c+a}{5}=\dfrac{a+b+b+c+c+a}{3+4+5}=\dfrac{2\left(a+b+c\right)}{12}=\dfrac{a+b+c}{6}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a+b}{3}=\dfrac{a+b+c}{6}\\\dfrac{b+c}{4}=\dfrac{a+b+c}{6}\\\dfrac{c+a}{5}=\dfrac{a+b+c}{6}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}3a+3b=3c\\2b+2c=4a\\a+c=b\end{matrix}\right.\)\(\Rightarrow a=b=c=0\)