Given $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$ for $\mathrm{△}ABC$, if $\frac{\cos A}{\alpha }=\frac{\cos B}{\beta }=\frac{\cos C}{\gamma }$ then $(\alpha ,\beta ,\gamma )$

$$\begin{gathered} b+c=11, c+a=12 \\ a+b=13 ; a+b+c=18 \\ a=7, b=6, c=5 \\ \cos A=\frac{b^2+c^2-a^2}{2bc}\Rightarrow =\frac{6^2+5^2-7^2}{2\times 6\times 5}=\frac{1}{5} \\ \cos b=\frac{c^2+a^2-b^2}{2ca}\Rightarrow =\frac{5^2+7^2-6^2}{2\times 7\times 5}=\frac{19}{35} \\ \cos C=\frac{a^2+b^2+c^2}{2ab}\Rightarrow =\frac{7^2+6^2-5^2}{2\times 7\times 6}=\frac{5}{7} \\ \cos A:\cos B:\cos C=\frac{1}{5}:\frac{19}{35}:\frac{5}{7}=7:19:25 \\ \frac{\cos A}{\alpha }=\frac{\cos \beta }{\beta }=\frac{\cos C}{\gamma } \\ \therefore (\alpha ,\beta ,\gamma )=(7,19,25) \end{gathered}$$