Binomial Theorem

If $26\left(\frac{2^3}{3}\left({ }^{12} \mathrm{C}_2\right)+\frac{2^5}{5}\left({ }^{12} \mathrm{C}_4\right)+\frac{2^7}{7}\left({ }^{12} \mathrm{C}_6\right)+\cdots+\frac{2^{13}}{13}\left({ }^{12} \mathrm{C}_{12}\right)\right)=3^{13}-\alpha$, then $\alpha$ is equal to :