If the line, 2 x-y+3=0 is at a distance

Question:

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

Solution:

$L_{1}: 2 x-y+3=0$

$L_{1}: 4 x-2 y+\alpha=0 \Rightarrow 2 x-y+\frac{\alpha}{2}=0$

$L_{1}: 6 x-3 y+\beta=0 \Rightarrow 2 x-y+\frac{\beta}{3}=0$

Distance between $L_{1}$ and $L_{2}$;

$\left|\frac{\alpha-6}{2 \sqrt{5}}\right|=\frac{1}{\sqrt{5}} \Rightarrow|\alpha-6|=2$

$\Rightarrow \alpha=4,8$

Distance between $L_{1}$ and $L_{3}$ :

$\left|\frac{\beta-9}{3 \sqrt{5}}\right|=\frac{2}{\sqrt{5}} \Rightarrow|\beta-9|=6$

$\Rightarrow \beta=15,3$

Sum of all values $=4+8+15+3=30$.