2 $$\times$$ (i) $$-$$ (ii) $$-$$ (iii) gives :

$$-$$ (1 + $$\beta$$)z = 3 $$-$$ $$\alpha$$

For infinitely many solution

$$\beta$$ + 1 = 0 = 3 $$-$$ $$\alpha$$ $$\Rightarrow$$ ($$\alpha$$, $$\beta$$) = (3, $$-$$1)

Hence, $$\alpha$$ + $$\beta$$ $$-$$ $$\alpha$$$$\beta$$ = 5