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:

$$z(\bar{z} + 2 + i) + k(2 + 3i) = 0$$

$$(x + iy)((x - iy) + 2 + i) + 2k + 3ki = 0$$

$$(x + iy)((x + 2) + i(1 - y)) + 2k + 3ki = 0$$

$$x(x + 2) + ix(1 - y) + iy(x + 2) + i^2y(1 - y) + 2k + 3ki = 0$$

$$\left(x^2 + 2x + y^2 - y + 2k\right) + i\left(x - xy + xy + 2y + 3k\right) = 0$$

$$\text{Real part: } x^2 + y^2 + 2x - y + 2k = 0 \quad \text{--- (Equation 1)}$$

$$\text{Imaginary part: } x + 2y + 3k = 0 \implies x = -(2y + 3k) \quad \text{--- (Equation 2)}$$

$$(-(2y + 3k))^2 + y^2 + 2(-(2y + 3k)) - y + 2k = 0$$

$$(4y^2 + 12ky + 9k^2) + y^2 - 4y - 6k - y + 2k = 0$$

$$5y^2 + y(12k - 5) + (9k^2 - 4k) = 0$$

$$D = B^2 - 4AC \ge 0$$ (Real Roots)

$$(12k - 5)^2 - 4(5)(9k^2 - 4k) \ge 0$$

$$-36k^2 - 40k + 25 \ge 0$$

$$36k^2 + 40k - 25 \le 0$$

$$9(\alpha + \beta) = 9 \left(-\frac{40}{36}\right) = -\frac{40}{4} = -10$$