Let the set of all values of $$k \in \mathbb{R}$$ such that the equation $$z(\bar{z} + 2 + i) + k(2 + 3i) = 0$$, $$z \in \mathbb{C}$$, has at least one solution, be the interval $$[\alpha, \beta]$$. Then $$9(\alpha + \beta)$$ is equal to:

A

$$-10$$

B

$$-8$$

C

$$10\sqrt{13}$$

D

$$8\sqrt{13}$$