If $$26\left(\frac{2^3}{3} {^{12} C_{2}} + \frac{2^5}{5} {^{12} C_{4}} + \frac{2^7}{7} {^{12} C_{6}} + \cdots + \frac{2^{13}}{13} {^{12} C_{12}}\right) = 3^{13} - \alpha$$, then $$\alpha$$ is equal to :

$$\frac{1}{k} \, ^nC_{k-1} = \frac{1}{n+1} \, ^{n+1}C_k$$

$$(1+x)^m = \sum_{k=0}^{m} \, ^mC_k x^k$$

$$S = \sum_{k=1}^{6} \frac{2^{2k+1}}{2k+1} \, ^{12}C_{2k}$$

$$S = \frac{1}{13} \sum_{k=1}^{6} 2^{2k+1} \, ^{13}C_{2k+1}$$ ($$\frac{1}{2k+1} \, ^{12}C_{2k} = \frac{1}{13} \, ^{13}C_{2k+1}$$)

$$S = \frac{2}{13} \sum_{k=1}^{6} 2^{2k} \, ^{13}C_{2k+1} = \frac{2}{13} \sum_{k=1}^{6} (2^2)^k \, ^{13}C_{2k+1}$$

$$S = \frac{2}{13} \sum_{k=1}^{6} 4^k \, ^{13}C_{2k+1}$$

Now, consider the expansion $$(1+x)^{13} = \, ^{13}C_0 + \, ^{13}C_1 x + \, ^{13}C_2 x^2 + \dots + \, ^{13}C_{13} x^{13}$$.

For $$x=2$$: $$(1+2)^{13} = 3^{13} = \sum_{r=0}^{13} \, ^{13}C_r 2^r$$

For $$x=-2$$: $$(1-2)^{13} = -1 = \sum_{r=0}^{13} \, ^{13}C_r (-2)^r$$

Subtracting the two:

$$3^{13} - (-1) = \sum_{r=0}^{13} \, ^{13}C_r (2^r - (-2)^r) = 2 \sum_{k=0}^{6} \, ^{13}C_{2k+1} 2^{2k+1}$$

$$3^{13} + 1 = 2 \left( \, ^{13}C_1 2^1 + \sum_{k=1}^{6} \, ^{13}C_{2k+1} 2^{2k+1} \right)$$

$$3^{13} + 1 = 2(13 \cdot 2 + 2 \sum_{k=1}^{6} \, ^{13}C_{2k+1} 4^k)$$

$$3^{13} + 1 = 52 + 4 \sum_{k=1}^{6} \, ^{13}C_{2k+1} 4^k$$

$$3^{13} - 51 = 4 \sum_{k=1}^{6} \, ^{13}C_{2k+1} 4^k$$

Since $$S = \frac{2}{13} \sum_{k=1}^{6} 4^k \, ^{13}C_{2k+1}$$, then $$13S = 2 \sum_{k=1}^{6} 4^k \, ^{13}C_{2k+1}$$:

$$26S = 4 \sum_{k=1}^{6} 4^k \, ^{13}C_{2k+1} = 3^{13} - 51$$

The original expression is $$26S = 3^{13} - \alpha$$. Comparing this to our derived $$26S = 3^{13} - 51$$, we find:

$$\alpha = 51$$