smc

Let $A$,$B$,$C$ denote the number of hours needed by Alpha, Beta, Gamma, respectively. We also have their respective efficiency $\frac{1}{A}$, $\frac{1}{B}$, and $\frac{1}{C}$. Thus we get the equations $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{A-6}$$ $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{B-1}$$ $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{2}{C}⟹\frac{1}{A}+\frac{1}{B}=\frac{1}{C}$$ Equating the first $2$ equations gets $$\frac{1}{A-6}=\frac{1}{B-1}⟹A=B+5$$ Substituting the new relation along with the third equation into the first equation gets $$\frac{2}{A}+\frac{2}{A-5}=\frac{1}{A-6}$$ Solving the quadratic gets $A=3$ or $\frac{20}{3}$. Since $B=A-5>0$, $A=\frac{20}{3}$ is the only legit solution. Thus $B=\frac{5}{3}$ and $h=\frac{1}{\frac{1}{A}+\frac{1}{B}}=\frac{4}{3}$, $\boxed{\frac{4}{3}}$.