Suppose you have the three numbers $2m$, $2m+3$ and $2m+5$. Each round you have to reduce every number by one. Before each round you can add half of some even number of your $3$ numbers to any of the other two remaining numbers.
You can do this rearranging as many times as you want before each round. Is it possible that after some round, two of the numbers are $0$ while the other is $> 0$?
Example: For $m=5$ you get $10,13,15$. The only even number is $10$ so you can add half of it to $13$ or $15$ so you would get $5,18,15$ or $5,13,20$. Now you could do the same with $18$ or $20$ respectively. Or you don‘t make any rearrangements at all so you get in the next round by subtracting 1 from every number to get $9,12,14$. You are free to do or not do as many possible rearrangements as you like and then get in the next round, as long as every number is at least 1 so you can subtract 1 from it.
With $m=1$:
Start with $(2,5,7)$. We get:
$$(2,5,7)\,{\mapsto}\, (1,4,6)\,\Rightarrow_3^1 (4,4,3)\mapsto \,(3,3,2)\Rightarrow_3^1 (4,3,1)\Rightarrow_1^3 (2,3,3)\mapsto $$
$$\mapsto(1,2,2)\Rightarrow_2^3 (1,1,3)\mapsto (0,0,2)$$
Where $\mapsto$ means that we subtract $1$ from each term and $\Rightarrow_i^j$ means that we take half of what is in cell $i$ and move it to cell $j$.