If the line, 2   x - y + 3 = 0 is at a distance 1 5 and 2 5 from the lines 4 x - 2 y + α = 0 and 6…

$2 x-y+3=0\dots \dots \left(i\right)$

$4x-2y+\alpha =0 \Rightarrow 2x-y+\frac{\alpha }{2}=0\dots \dots \left(ii\right)$$6x-3y+\beta =0 \Rightarrow 2x-y+\frac{\beta }{3}=0\dots \dots \left(iii\right)$

${\mathrm{d}}_1=\frac{\left|\frac{\alpha }{2}-3\right|}{\sqrt{2^2+1^2}}=\frac{1}{\sqrt{5}}\Rightarrow |\alpha -6|=2$

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

${\mathrm{d}}_2=\frac{\left|\frac{\beta }{3}-3\right|}{\sqrt{2^2+1^2}}=\frac{2}{\sqrt{5}} \Rightarrow |\beta -9|=6$

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

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