1. Given lines

The reference line is $2x-y+3=0.$

The other two lines are $4x-2y+\alpha=0$ and $6x-3y+\beta=0.$

Notice that all three lines are parallel, since:

• $4x-2y+\alpha=0$ has slope $2$
• $6x-3y+\beta=0$ has slope $2$
• $2x-y+3=0$ also has slope $2$

So we can use the distance formula for parallel lines.

2. Distance between $2x-y+3=0$ and $4x-2y+\alpha=0$

First divide the second line by $2$: $2x-y+\frac{\alpha}{2}=0.$

Now compare with $2x-y+3=0.$

For parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$, distance is $d=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}.$

Thus, $\frac{\left|3-\frac{\alpha}{2}\right|}{\sqrt{2^2+(-1)^2}}=\frac{1}{\sqrt{5}}.$

Since $\sqrt{2^2+(-1)^2}=\sqrt{5}$, $\frac{\left|3-\frac{\alpha}{2}\right|}{\sqrt{5}}=\frac{1}{\sqrt{5}}.$

So, $\left|3-\frac{\alpha}{2}\right|=1.$

Hence, $3-\frac{\alpha}{2}=\pm 1.$

Case 1: $3-\frac{\alpha}{2}=1 \implies \frac{\alpha}{2}=2 \implies \alpha=4.$

Case 2: $3-\frac{\alpha}{2}=-1 \implies \frac{\alpha}{2}=4 \implies \alpha=8.$

So possible values of $\alpha$ are: $\alpha=4,\ 8.$

3. Distance between $2x-y+3=0$ and $6x-3y+\beta=0$

Divide the second line by $3$: $2x-y+\frac{\beta}{3}=0.$

Now distance from $2x-y+3=0$ is $\frac{\left|3-\frac{\beta}{3}\right|}{\sqrt{5}}=\frac{2}{\sqrt{5}}.$

Therefore, $\left|3-\frac{\beta}{3}\right|=2.$

So, $3-\frac{\beta}{3}=\pm 2.$

Case 1: $3-\frac{\beta}{3}=2 \implies \frac{\beta}{3}=1 \implies \beta=3.$

Case 2: $3-\frac{\beta}{3}=-2 \implies \frac{\beta}{3}=5 \implies \beta=15.$

So possible values of $\beta$ are: $\beta=3,\ 15.$

4. Sum of all possible values of $\alpha$ and $\beta$

Sum of possible $\alpha$ values: $4+8=12.$

Sum of possible $\beta$ values: $3+15=18.$

Therefore total sum is $12+18=30.$

5. Final answer

$\boxed{30}$

The derived answer matches the stored correct answer.