If the line, 2x - y + 3 = 0 is at a distance 1 5 and 2 5 from the...

$2x - y + 3 = 0 \ldots \ldots ( i )$

$4x - 2y + \alpha = 0 ⇒ 2x - y + \dfrac{\alpha}{2} = 0 \ldots \ldots ( ii )$$6x - 3y + \beta = 0 ⇒ 2x - y + \dfrac{\beta}{3} = 0 \ldots \ldots ( iii )$

${d}_{1} = \dfrac{\left|\dfrac{\alpha}{2} - 3\right|}{\sqrt{{2}^{2} + {1}^{2}}} = \dfrac{1}{\sqrt{5}} ⇒ | \alpha - 6 | = 2$

$⇒ \alpha - 6 = 2, - 2 ⇒ \alpha = 8, 4$

${d}_{2} = \dfrac{\left|\dfrac{\beta}{3} - 3\right|}{\sqrt{{2}^{2} + {1}^{2}}} = \dfrac{2}{\sqrt{5}} ⇒ | \beta - 9 | = 6$

$⇒ \beta - 9 = 6, - 6 ⇒ \beta = 15, 3$

Sum of all values of $\alpha$ and $\beta = 30$.