\(19,\dfrac{a+b}{6}=\dfrac{b+c}{7}=\dfrac{c+a}{8}=\dfrac{2\left(a+b+c\right)}{6+7+8}=\dfrac{28}{21}=\dfrac{4}{3}\\ \Rightarrow\left\{{}\begin{matrix}a+b=8\\b+c=\dfrac{28}{3}\\c+a=\dfrac{32}{3}\end{matrix}\right.\\ \Rightarrow c+a=c+8-b=c+8-\dfrac{28}{3}+c=\dfrac{32}{3}\\ \Rightarrow c=6\left(A\right)\)