If the line $2x-y+3=0$ is at distances $\frac{1}{\sqrt{5}}\text{ and }$ from the lines $4x-2y+\alpha =0\text{ and }6x-3y+\beta =0$ respectively the sum of all possible values of a and $\beta ,$, is

It is given that

$\begin{array}{l}\left|\frac{3-\alpha }{2}\right|=\frac{1}{\sqrt{5}}\text{ and }\frac{\left|3-\frac{\beta }{3}\right|}{\sqrt{4+1}}=\frac{2}{\sqrt{5}}\\ \Rightarrow \left|3-\frac{\alpha }{2}\right|=1\text{ and }\left|3-\frac{\beta }{3}\right|=2\\ \Rightarrow |6-\alpha |=2\text{ and }|9-\beta |=6\end{array}$

$\begin{array}{l}\Rightarrow 6-\alpha =\pm 2\text{ and }9-\beta =\pm 6\\ \Rightarrow \alpha =4,8\text{ and }\beta =3,15\end{array}$

Hence, the sum of all possible values of $\alpha \text{ and }\beta$ is 4 + 8 + 3 + 15 = 30 .