Look for a tinge of familiar

$\begin{aligned} S & = \sum_{n=1}^\infty \frac{(-1)^{n+1}(5n+24)}{n^2} \\ & = 5 \color{#3D99F6} {\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}} + 24 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2} \quad \quad \small \color{#3D99F6}{\text{As } \ln (1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}} \\ & = 5\color{#3D99F6}{\ln 2} + 24\left(\color{#D61F06}{ \sum_ {n=1}^\infty \frac{1}{n^2}} - \frac{2}{2^2} \color{#D61F06}{ \sum_ {n=1}^\infty \frac{1}{n^2}} \right) \quad \quad \small \color{#D61F06}{\text{Basel problem: }\sum_ {n=1}^\infty \frac{1}{n^2} = \zeta(2) = \frac{\pi^2}{6}} \\ & = 5 \ln 2 + 12 \left(\color{#D61F06}{\frac{\pi^2}{6}}\right) \\ & = 5 \ln 2 + 2\pi^2 \\ & = \ln \left(32e^{2\pi^2}\right) \end{aligned}$

$\Rightarrow A+B+C = 32+2+2 = \boxed{36}$