$\displaystyle S = \sum_{r = 1}^{\infty } \frac{ r + 2}{2^{r+1}.r(r + 1)}$

$\displaystyle = \sum_{r = 1}^{\infty} \frac{2(r+1) -r}{2^{r+1}.r(r+1)}$

Now seperating

$\displaystyle \sum_{r=1}^{\infty } \frac{1}{2^{r + 1}}( \frac{2}{r} - \frac{1}{r + 1})$

$\displaystyle = \sum_{r = 1}^{\infty} ( \frac{1}{2^{r}.r} - \frac{1}{2^{r+1}.(r+1)})$

All the terms will cancel out as we substitute values of 'r' (first term will remain as it is)

Therefore we finally get

S = $\frac{1}{2}$

Hence

$\displaystyle S^{-12} = \boxed{4096}$