Mathematics Question from JEE Main 2024 · 29 January, Shift 2

Let $\alpha$, $\beta$ be the roots of the equation $x^2 - \sqrt{6}x + 3 = 0$ such that $\operatorname{Im}(\alpha) > \operatorname{Im}(\beta)$. 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: