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

We have the given line $$2x - y + 3 = 0$$ and two other lines $$4x - 2y + \alpha = 0$$ and $$6x - 3y + \beta = 0$$ which are both obviously parallel to the first line because their left-hand sides are constant multiples of $$2x - y$$. For parallel lines we use the formula for the perpendicular distance between the pair of lines

$$\text{Distance} \;=\; \frac{\lvert C_1 - C_2\rvert}{\sqrt{A^{2} + B^{2}}},$$

where the lines are written in the common normal form $$Ax + By + C_1 = 0$$ and $$Ax + By + C_2 = 0$$.

First we compare the first line with $$4x - 2y + \alpha = 0$$. To make the two equations share the same normal vector, we divide the entire second equation by $$2$$:

$$\frac{4x - 2y + \alpha}{2} = 0 \;\;\Longrightarrow\;\; 2x - y + \frac{\alpha}{2} = 0.$$

Now both lines are in the form $$2x - y + C = 0$$, so we can directly apply the distance formula. The required distance is given to be $$\dfrac{1}{\sqrt{5}}$$, and the denominator $$\sqrt{A^{2} + B^{2}}$$ equals $$\sqrt{2^{2} + (-1)^{2}} = \sqrt{5}$$. Substituting these into the formula we get

$$\frac{\lvert 3 - \dfrac{\alpha}{2}\rvert}{\sqrt{5}} = \frac{1}{\sqrt{5}} \;\;\Longrightarrow\;\; \lvert 3 - \dfrac{\alpha}{2}\rvert = 1.$$

This absolute-value equation splits into two linear equations:

1. $$3 - \dfrac{\alpha}{2} = 1 \;\;\Longrightarrow\;\; \dfrac{\alpha}{2} = 2 \;\;\Longrightarrow\;\; \alpha = 4,$$

2. $$3 - \dfrac{\alpha}{2} = -1 \;\;\Longrightarrow\;\; \dfrac{\alpha}{2} = 4 \;\;\Longrightarrow\;\; \alpha = 8.$$

Hence the possible values of $$\alpha$$ are $$4$$ and $$8$$.

Next we compare $$2x - y + 3 = 0$$ with the line $$6x - 3y + \beta = 0$$. We again normalise by dividing by $$3$$:

$$\frac{6x - 3y + \beta}{3} = 0 \;\;\Longrightarrow\;\; 2x - y + \frac{\beta}{3} = 0.$$

The distance between these two parallel lines is stated to be $$\dfrac{2}{\sqrt{5}}$$. Using the same denominator $$\sqrt{5}$$, we write

$$\frac{\lvert 3 - \dfrac{\beta}{3}\rvert}{\sqrt{5}} = \frac{2}{\sqrt{5}} \;\;\Longrightarrow\;\; \lvert 3 - \dfrac{\beta}{3}\rvert = 2.$$

Again we split the absolute value:

1. $$3 - \dfrac{\beta}{3} = 2 \;\;\Longrightarrow\;\; \dfrac{\beta}{3} = 1 \;\;\Longrightarrow\;\; \beta = 3,$$

2. $$3 - \dfrac{\beta}{3} = -2 \;\;\Longrightarrow\;\; \dfrac{\beta}{3} = 5 \;\;\Longrightarrow\;\; \beta = 15.$$

Thus the possible values of $$\beta$$ are $$3$$ and $$15$$.

The problem asks for the sum of all possible values of $$\alpha$$ and $$\beta$$ together. Adding these distinct values we get

$$\underbrace{4 + 8}_{\alpha\text{-values}} + \underbrace{3 + 15}_{\beta\text{-values}} = 12 + 18 = 30.$$

Hence, the correct answer is 30.