If $2x+3y+4=0$  is the perpendicular bisector of the segment joining the points $A\left(1,2\right)$  and  $B\left(\alpha ,\beta \right)$ , then the value of $\alpha +\beta$  is

$2x+3y+4=0$

$\Rightarrow$  slope $\left(m_1\right)=\frac{-2}{3}$

Point $C=$  midpoint of $AB$

$=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

$=\left(\frac{1+\alpha }{2},\frac{2+\beta }{2}\right)$

$C$  is line on $2x+3y+4=0$

$2\left(\frac{1+\alpha }{2}\right)+3\left(\frac{2+\beta }{2}\right)+4=0$

$1+\alpha +\frac{6+3\beta }{2}+4=0$

$\frac{2\alpha +6+3\beta +10}{2}=0$

$2\alpha +3\beta +16=0\to \left(1\right)$

Slope of $AB=\frac{3}{2}$

$\frac{y_2-y_1}{x_2-x_1}=\frac{3}{2}$

$\frac{\beta -2}{\alpha -1}=\frac{3}{2}$

$2\beta -4=3\alpha -3$

$3\alpha -2\beta +1=0\to \left(2\right)$

$\begin{array}{c}2\times \left(1\right)\Rightarrow 4\alpha +6\beta +32=0\\ \underset{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}{3\times \left(2\right)\Rightarrow 9\alpha -6\beta +3=0}\\ 13\alpha +35=0\end{array}$

$\alpha =\frac{-35}{13}$

From $\left(2\right)$

$3\left(\frac{-35}{13}\right)-2\beta +1=0$

$2\beta =\frac{-105}{13}+1$

$2\beta =\frac{-92}{13}$

$\Rightarrow \beta =\frac{-46}{13}$

$\alpha +\beta =\frac{-35}{13}-\frac{46}{13}$

$=\frac{-81}{13}$