Solution
Put $\mathrm{z}=\mathrm{x}+\mathrm{iy}, \overline{\mathrm{z}}=\mathrm{x}-\mathrm{iy}$
\[
\begin{aligned}
& \mathrm{z}(\overline{\mathrm{z}}+2+\mathrm{i})+\mathrm{k}(2+3 \mathrm{i})=0 \\
& \Rightarrow(\mathrm{x}+\mathrm{iy})(\mathrm{x}-\mathrm{iy}+2+\mathrm{i})+2 \mathrm{k}+3 \mathrm{ki}=0 \\
& \Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{x}-\mathrm{y}+2 \mathrm{k}+(\mathrm{x}+2 \mathrm{y}+3 \mathrm{k}) \mathrm{i}=0+\mathrm{i} 0 \\
& \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{x}-\mathrm{y}+2 \mathrm{k}=0 \text { and } \mathrm{x}+2 \mathrm{y}+3 \mathrm{k}=0 \Rightarrow \mathrm{x}=-2 \mathrm{y}-3 \mathrm{k}
\end{aligned}
\]
eliminating x from first equation
\[
\begin{aligned}
& 5 \mathrm{y}^{2}+(12 \mathrm{k}-5) \mathrm{y}+9 \mathrm{k}^{2}-4 \mathrm{k}=0 \text { since } \mathrm{y} \in \mathrm{R} \text { so } \mathrm{D} \geq 0 \\
& (12 \mathrm{k}-5)^{2}-4.5\left(9 \mathrm{k}^{2}-4 \mathrm{k}\right) \geq 0 \Rightarrow 36 \mathrm{k}^{2}+40 \mathrm{k}-25 \leq 0 \text { then } \mathrm{k} \in[\alpha, \beta] \\
& \Rightarrow \alpha+\beta=-\frac{10}{9} \Rightarrow 9(\alpha+\beta)=-10
\end{aligned}
\]
.
