Let the set of all values of k ∈ ℝ such that the equation z( z + 2...

Let $z = x + iy$, then $\bar{z} = x - iy$.

Substituting $z$ into the given equation:

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

$x^2 + y^2 + 2x + ix + 2iy - y + 2k + 3ki = 0$

Separating the real and imaginary parts, we get:

Real part: $x^2 + y^2 + 2x - y + 2k = 0$

Imaginary part: $x + 2y + 3k = 0 \Rightarrow x = -2y - 3k$

Substituting $x$ into the real part equation:

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

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

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

For the equation to have at least one solution $z \in \mathbb{C}$, there must be at least one real value of $y$. Thus, the discriminant of this quadratic equation in $y$ must be non-negative ($\Delta \ge 0$):

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

$144k^2 - 120k + 25 - 180k^2 + 80k \ge 0$

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

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

The values of $k$ lie in the interval $[\alpha, \beta]$, where $\alpha$ and $\beta$ are the roots of the equation $36k^2 + 40k - 25 = 0$.

The sum of the roots is given by:

$\alpha + \beta = -\dfrac{40}{36} = -\dfrac{10}{9}$

Therefore, $9(\alpha + \beta) = 9 \left( -\dfrac{10}{9} \right) = -10$.

Answer: $-10$