Babai wrote: Find all positive integer solutions $x,y,z$ such that $1/x +2/y - 3/z=1$ I found some roots: 1) $x=1$ and $z=3/2y$ (with $y=2k, \; \forall k=1,2,...$) 2) $y=2,z=3x$

I have understood what you want to say.But there may be other solutions so please give all possible solutions.

Babai wrote: Find all positive integer solutions $x,y,z$ such that $1/x +2/y - 3/z=1$ 1) $x=1$ and the equation is $\frac 2y=\frac 3z$ and the solutions $(x,y,z)=(1,2n,3n)$ 2) $x>1$ and $y=1$ and the equaiton is $\frac 1x+1=\frac 3z$ and so $\frac 3z>1$ and $z<3$ and : 2.1) $z=1$ and the equaiton is $\frac 1x=2$ and no solution 2.2) $z=2$ and the equaiton is $\frac 1x+1=\frac 32$ and the solutions $(x,y,z)=(2,1,2)$ 3) $x>1$ and $y=2$ and the equation is $\frac 1x=\frac 3z$ and the solutions $(x,y,z)=(n,2,3n)$ 4) $x>1$ and $y>2$ $\frac 1x+\frac 2y=1+\frac 3z>1$ and so $\frac 2y>1-\frac 1x >1-\frac 12$ and so $y<4$ and so $y=3$ $\frac 1x+\frac 23=1+\frac 3z>1$ and so $\frac 1x > \frac 13$ and so $x<3$ and so $x=2$ And the equation is $\frac 12+\frac 23-\frac 3z=1$ and so $z=18$ and the solution $(x,y,z)=(2,3,18)$ Hence the solutions : $(2,1,2)$ $(2,3,18)$ $(1,2n,3n)$ $\forall n\in\mathbb N$ $(n,2,3n)$ $\forall n\in\mathbb N$