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(B) The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.For the

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$30$

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(B) The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.For the first pair,rewrite $2x - y + 3 = 0$ as $4x - 2y + 6 = 0$. The distance is $\frac{1}{\sqrt{5}} = \frac{|\alpha - 6|}{\sqrt{4^2 + (-2)^2}} = \frac{|\alpha - 6|}{\sqrt{20}} = \frac{|\alpha - 6|}{2\sqrt{5}}$.Thus,$|\alpha - 6| = 2$,which gives $\alpha - 6 = 2$ or $\alpha - 6 = -2$,so $\alpha = 8$ or $\alpha = 4$.For the second pair,rewrite $2x - y + 3 = 0$ as $6x - 3y + 9 = 0$. The distance is $\frac{2}{\sqrt{5}} = \frac{|\beta - 9|}{\sqrt{6^2 + (-3)^2}} = \frac{|\beta - 9|}{\sqrt{45}} = \frac{|\beta - 9|}{3\sqrt{5}}$.Thus,$|\beta - 9| = 6$,which gives $\beta - 9 = 6$ or $\beta - 9 = -6$,so $\beta = 15$ or $\beta = 3$.The sum of all possible values of $\alpha$ and $\beta$ is $(8 + 4) + (15 + 3) = 12 + 18 = 30$.

If the line $2x - y + 3 = 0$ is at a distance of $\frac{1}{\sqrt{5}}$ 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:

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