Let $$H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$, $$a > 0$$, $$b > 0$$, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $$4(2\sqrt{2} + \sqrt{14})$$. If the eccentricity $$H$$ is $$\frac{\sqrt{11}}{2}$$, then value of $$a^2 + b^2$$ is equal to ______.

Given hyperbola $$H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ with $$a > 0, b > 0$$, its transverse axis length is $$2a$$ and its conjugate axis length is $$2b$$, and their sum is $$2a + 2b = 4(2\sqrt{2} + \sqrt{14}),$$ so that $$a + b = 2(2\sqrt{2} + \sqrt{14}) = 4\sqrt{2} + 2\sqrt{14}.$$

The eccentricity is $$e = \frac{\sqrt{11}}{2},$$ hence $$e^2 = \frac{11}{4} = 1 + \frac{b^2}{a^2},$$ giving $$\frac{b^2}{a^2} = \frac{7}{4}, \quad b = \frac{a\sqrt{7}}{2}.$$

Substituting into the sum equation yields $$a + \frac{a\sqrt{7}}{2} = 4\sqrt{2} + 2\sqrt{14},$$ or equivalently $$a\left(1 + \frac{\sqrt{7}}{2}\right) = 4\sqrt{2} + 2\sqrt{14},$$ so that $$a \cdot \frac{2 + \sqrt{7}}{2} = 4\sqrt{2} + 2\sqrt{14}.$$ Noting that $$4\sqrt{2} + 2\sqrt{14} = 2\sqrt{2}(2 + \sqrt{7}),$$ we find $$a = \frac{2 \cdot 2\sqrt{2}(2 + \sqrt{7})}{2 + \sqrt{7}} = 4\sqrt{2}.$$

It follows that $$b = \frac{4\sqrt{2} \cdot \sqrt{7}}{2} = 2\sqrt{14},$$ and hence $$a^2 = 32, \quad b^2 = 4 \times 14 = 56, \quad a^2 + b^2 = 32 + 56 = 88.$$

The answer is $$\boxed{88}$$.