If a, b, and c are the sides of a$∆ABC$ such that  $\frac{a+b}{11}=\frac{b+c}{13}=\frac{c+a}{12}$ and $\frac{\cos A}{p}=\frac{\cos B}{q}=\frac{\cos C}{r}$, then

We have, $\frac{a+b}{11}=\frac{b+c}{13}=\frac{c+a}{12}=\lambda$ (let)

Then, $a+b=11\lambda$    ... (i)

$\begin{array}{l}\mathrm{b}+\mathrm{c}=13\mathrm{\lambda } \end{array}$ (ii)

$\mathrm{c}+\mathrm{a}=12\mathrm{\lambda }$ ...(iii)

On adding Eqs. (i), (ii) and (iii)1 we get

$\begin{array}{l}2(\mathrm{a}+\mathrm{b}+\mathrm{c})=36\mathrm{\lambda }\\ \mathrm{a}+\mathrm{b}+\mathrm{c}=18\mathrm{\lambda } ... \left(\mathrm{iv}\right)\end{array}$

On solving Eqs. (i) and (iv), (ii) and (iv)1 (Iii) and (iv), we get

$c=7\lambda ,a=5\lambda ,b=6\lambda$

Now,  $\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\begin{array}{l}=\frac{36{\mathrm{\lambda }}^2+49{\mathrm{\lambda }}^2-25{\mathrm{\lambda }}^2}{2\times 6\mathrm{\lambda }\times 7\mathrm{\lambda }}\\ =\frac{36+49-25}{2\times 7\times 6}=\frac{5}{7}\\ \end{array}$

$\therefore \frac{\cos \mathrm{A}}{\frac{5}{7}}=1$ ...(v)

Similarly, $\frac{\cos B}{19/35}=1$ and $\frac{\cos C}{1/5}=1$

$\begin{array}{l}\therefore \frac{\cos A}{5/7}=\frac{\cos B}{19/35}=\frac{\cos C}{1/5}\\ \frac{\cos A}{25}=\frac{\cos B}{19}=\frac{\cos C}{7}\end{array}$

But given that $\frac{\cos A}{p}=\frac{\cos B}{q}=\frac{\cos C}{r}$

$\therefore p=25,q=19,r=7$