If $a>=7$ then $1/a+2/b+3/c <= 1/7+2/7+3/7=6/7<1$, so $a=2, 3$ or $5$. 1) $a=2$ therefore $4/b+6/c=1$ so $bc-6b-4c=0$ or $(b-4)(c-6)=24$. If $b>=11$ then $4/b+6/c<=10/11<1$ so $5<=b<=10$. We get the solutions $(b, c) = (5, 30), (6, 18), (7, 14), (8, 12)$ and $(10, 10)$. 2) $a=3$ $=>$ $6/b+9/c=2$. Obviously $b>=4$ and if $b>=16$ then $6/b+9/c<=15/16<1$ so $4<=b<=15$. We rearrange to get $2bc-9b-6c=0$ or $(2c-9)(b-3)=27$. For $(b, c)$ we get $(4, 18)$ and $(6, 9)$. 3)$a=5$ so $10/b+15/c=4$. It is easy to see that $b>=3$. Again $8bc-30b-20c=0$ or $(2b-5)(4c-15)=75$. Here for $(b, c)$ we get $(4, 10)$.

@above Since $a \le b \le c$ $\implies$ $a=5$ has no solutions!

Night_Witch123 wrote: @above Since $a \le b \le c$ $\implies$ $a=5$ has no solutions! We also get a solution for $a=b=c=6$

HoRI_DA_GRe8 wrote: Night_Witch123 wrote: @above Since $a \le b \le c$ $\implies$ $a=5$ has no solutions! We also get a solution for $a=b=c=6$ $a $ is given to be prime...