Romanian Mathematical Olympiad

We rewrite the given equality as: $$(x-2{)}^2+(2y-3{)}^2+(3z-4{)}^2=9.$$ This implies $(x-2{)}^2\le 9$, $(2y-3{)}^2\le 9$ and $(3z-4{)}^2\le 9$, therefore $|x-2|\le 3$, $|2y-3|\le 3$, $|3z-4|\le 3$ and the last inequalities are equivalent to the conclusion.