It is equivalent to solving $$\begin{array}{rl} & C_n^k : C_n^{k+1} : C_n^{k+2} = 3: 4:5 \\ \\ \implies & \frac{n!}{k!(n-k)!} : \frac{n!}{(k+1)!(n-k-1)!} : \frac{n!}{(k+2)!(n-k-2)!}=3:4:5 \\ \implies & \left\{\begin{array}{rcl} (n-k) : (k+1) &=& 4:3 \\ (n-k-1):(k+2) &=& 5:4 \end{array}\right. \\ \\ \implies & n = \boxed{62}, k =26\end{array}$$