Given Equation

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

where

$k\in\mathbb{R}$

and $z\in\mathbb{C}$.

We need values of $k$ for which at least one solution exists.

Step 1 : Let

$z=x+iy$

Then

$\bar z=x-iy$

Now,

$z(\bar z+2+i)=(x+iy)(x-iy+2+i)$

Write

$\bar z+2+i=(x+2)+i(1-y)$

Hence,

$z(\bar z+2+i)=(x+iy)\big((x+2)+i(1-y)\big)$

Multiplying,

$=(x(x+2)-y(1-y))+i\big(x(1-y)+y(x+2)\big)$

Real part:

$x^2+2x-y+y^2$

Imaginary part:

$x-yx+yx+2y=x+2y$

Thus,

$z(\bar z+2+i)=(x^2+y^2+2x-y)+i(x+2y)$

Equation becomes

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

Equating real and imaginary parts:

$x^2+y^2+2x-y+2k=0$

$x+2y+3k=0$

Step 2 : Express $k$

From second equation,

$k=-\dfrac{x+2y}{3}$

Substitute into first equation:

$x^2+y^2+2x-y-\dfrac{2(x+2y)}{3}=0$

Multiply by $3$:

$3x^2+3y^2+6x-3y-2x-4y=0$

$3x^2+3y^2+4x-7y=0$

Complete squares:

$3\left(x^2+\dfrac43x\right)+3\left(y^2-\dfrac73y\right)=0$

$3\left[\left(x+\dfrac23\right)^2-\dfrac49\right]+3\left[\left(y-\dfrac76\right)^2-\dfrac{49}{36}\right]=0$

$3\left(x+\dfrac23\right)^2+3\left(y-\dfrac76\right)^2=\dfrac43+\dfrac{49}{12}$

$3\left(x+\dfrac23\right)^2+3\left(y-\dfrac76\right)^2=\dfrac{65}{12}$

$\left(x+\dfrac23\right)^2+\left(y-\dfrac76\right)^2=\dfrac{65}{36}$

So the circle has center

$\left(-\dfrac23,\dfrac76\right)$

and radius

$\dfrac{\sqrt{65}}{6}$

Step 3 : Find range of $k$

Since

$k=-\dfrac{x+2y}{3}$

we find range of $x+2y$.

For a circle, maximum and minimum of $x+2y$ are

$\left(-\dfrac23\right)+2\left(\dfrac76\right)\pm\dfrac{\sqrt{1^2+2^2}\sqrt{65}}{6}$

$=\dfrac53\pm\dfrac{\sqrt5\sqrt{65}}{6}$

$=\dfrac53\pm\dfrac{5\sqrt{13}}{6}$

Thus,

$x+2y\in\left[\dfrac53-\dfrac{5\sqrt{13}}{6},\dfrac53+\dfrac{5\sqrt{13}}{6}\right]$

Therefore,

$k\in\left[-\dfrac13\left(\dfrac53+\dfrac{5\sqrt{13}}{6}\right),-\dfrac13\left(\dfrac53-\dfrac{5\sqrt{13}}{6}\right)\right]$

Hence,

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

Therefore,

$9(\alpha+\beta)=-10$

Final Answer

$\boxed{-10}$