If the line, 2x -y+3=0 is at a distance $1/\sqrt{5}and2/\sqrt{5}$ from the lines $4x-2y+\alpha =0and6x-3y+\beta =0,$ respectively, then the sum of all possible values of $\alpha and\beta$ is …….

Distance between $2x-y+3=0,4x-2y+\alpha =0$ is $\frac{1}{\sqrt{5}}$

$\Rightarrow \left|\frac{{\displaystyle \alpha -6}}{{\displaystyle \sqrt{16+4}}}\right|=\frac{1}{\sqrt{5}}\Rightarrow |\alpha -6|=2\Rightarrow \alpha =8or4$

Distance between $2x-y+3=0,6x-3y+\beta =0is\frac{2}{\sqrt{5}}\Rightarrow \left|\frac{{\displaystyle \beta -9}}{{\displaystyle \sqrt{36+9}}}\right|=\frac{3}{10}$

$\Rightarrow |\beta -9|=6\Rightarrow \beta =15or3. \therefore \alpha +\beta =23or11or19or7.$

Their sum = 60