If $$1 + \frac{\sqrt{3} - \sqrt{2}}{2\sqrt{3}} + \frac{5 - 2\sqrt{6}}{18} + \frac{9\sqrt{3} - 11\sqrt{2}}{36\sqrt{3}} + \frac{49 - 20\sqrt{6}}{180} + \ldots$$ upto $$\infty = 2 + \left(\sqrt{\frac{b}{a}} + 1\right)\log_e\left(\frac{a}{b}\right)$$, where $$a$$ and $$b$$ are integers with $$\gcd(a, b) = 1$$, then $$11a + 18b$$ is equal to ______

Observe that $$\frac{\sqrt3-\sqrt2}{\sqrt3}= 1-\sqrt{\frac23}$$

Let $$y=1-\sqrt{\frac23}$$

Then,

$$\frac{\sqrt3-\sqrt2}{2\sqrt3}=\frac y2$$

Also,

$$y^2=\left(1-\sqrt{\frac23}\right)^2=\frac{5-2\sqrt6}{3}$$

Hence,

$$\frac{5-2\sqrt6}{18}=\frac{y^2}{6}$$

Similarly,

$$\frac{9\sqrt3-11\sqrt2}{36\sqrt3}=\frac{y^3}{12},\qquad\frac{49-20\sqrt6}{180}=\frac{y^4}{20}$$

Therefore,

$$S=1+\sum_{n=1}^{\infty}\frac{y^n}{n(n+1)}$$

Using

$$\frac1{n(n+1)}=\frac1n-\frac1{n+1},$$

we get

$$S=1+\sum_{n=1}^{\infty}\frac{y^n}{n}-\sum_{n=1}^{\infty}\frac{y^n}{n+1}$$

Now,

$$\sum_{n=1}^{\infty}\frac{y^n}{n}=-\ln(1-y)$$

and

$$\sum_{n=1}^{\infty}\frac{y^n}{n+1}=\frac1y\left(-\ln(1-y)-y\right)$$

Hence,

$$S=1-\ln(1-y)+\frac{\ln(1-y)}{y}+1$$

$$=2+\frac{1-y}{y}\ln(1-y)$$

Since

$$1-y=\sqrt{\frac23},$$

$$\frac{1-y}{y}=\frac{\sqrt{\frac23}}{1-\sqrt{\frac23}}$$

Rationalizing,

$$=\sqrt6+2$$

Thus,

$$S=2+(\sqrt6+2)\ln\left(\sqrt{\frac23}\right)$$

Using

$$\ln\left(\sqrt{\frac23}\right)=\frac12\ln\left(\frac23\right),$$

we get

$$S=2+\left(\frac{\sqrt6}{2}+1\right)\ln\left(\frac23\right)$$

But

$$\frac{\sqrt6}{2}=\sqrt{\frac32}$$

Hence,

$$S=2+\left(\sqrt{\frac32}+1\right)\ln\left(\frac23\right)$$

Comparing with

$$2+\left(\sqrt{\frac ba}+1\right)\ln\left(\frac ab\right),$$

we get

$$a=2,\qquad b=3$$

Therefore,

$$11a+18b=11(2)+18(3)$$

$$=22+54$$

$$=76$$

Hence, the required answer is

$$\boxed{76}$$