Let $$\alpha, \beta$$ be the roots of the equation $$x^2 - \sqrt{6}x + 3 = 0$$ such that $$\text{Im}(\alpha) > \text{Im}(\beta)$$. Let $$a, b$$ be integers not divisible by $$3$$ and $$n$$ be a natural number such that $$\frac{\alpha^{99}}{\beta} + \alpha^{98} = 3^n(a + ib), i = \sqrt{-1}$$. Then $$n + a + b$$ is equal to ______.

The roots of $$x^2 - \sqrt{6}x + 3 = 0$$ are:

$$ \alpha, \beta = \frac{\sqrt{6} \pm \sqrt{6 - 12}}{2} = \frac{\sqrt{6} \pm i\sqrt{6}}{2} = \frac{\sqrt{6}}{2}(1 \pm i) $$

Since $$\text{Im}(\alpha) > \text{Im}(\beta)$$: $$\alpha = \frac{\sqrt{6}}{2}(1 + i)$$, $$\beta = \frac{\sqrt{6}}{2}(1 - i)$$.

Convert to polar form: $$|\alpha| = |\beta| = \frac{\sqrt{6}}{2}\sqrt{2} = \sqrt{3}$$

$$\alpha = \sqrt{3}\,e^{i\pi/4}, \quad \beta = \sqrt{3}\,e^{-i\pi/4}$$

 $$\frac{\alpha}{\beta} = e^{i\pi/2} = i$$

 $$\alpha^{98} = (\sqrt{3})^{98} \cdot e^{i \cdot 98\pi/4} = 3^{49} \cdot e^{i \cdot 49\pi/2}$$

Now $$\frac{49\pi}{2} = 24\pi + \frac{\pi}{2}$$, so $$e^{i \cdot 49\pi/2} = e^{i\pi/2} = i$$.

Therefore $$\alpha^{98} = 3^{49} \cdot i$$.

 $$ \frac{\alpha^{99}}{\beta} + \alpha^{98} = \alpha^{98}\left(\frac{\alpha}{\beta} + 1\right) = 3^{49} \cdot i \cdot (i + 1) = 3^{49}(i + i^2) = 3^{49}(-1 + i) $$

Match with the given form: $$3^{49}(-1 + i) = 3^n(a + ib)$$

So $$n = 49$$, $$a = -1$$, $$b = 1$$. Both $$a$$ and $$b$$ are not divisible by 3. 

$$ n + a + b = 49 + (-1) + 1 = 49 $$