Damn irrationals

Note that by applying the operation, the sum of the squares of the three numbers remains constant. Let's prove this:

Let the numbers be $(a,b,c)$ . Thus, the new numbers are $(\frac{a+b}{\sqrt{2}}, \frac{a-b}{\sqrt{2}}, c)$ . Here, the sum of the squares of each number is

$\frac{(a+b)^2}{2}+\frac{(a-b)^2}{2}+c^2=\frac{2a^2+2b^2}{2}+c^2=a^2+b^2+c^2$

Therefore, the sum of the squares of all the numbers remains constant.

Note that $4^2+(1+2\sqrt{2})^2+\sqrt{2}^2 = 16+1+8+4\sqrt{2}+2=27+4\sqrt{2}$ , whereas $(3+\sqrt{2})^2+(2\sqrt{2})^2+(\sqrt{2}-1)^2=9+2+6\sqrt{2}+8+2+1-2\sqrt{2}=22+4\sqrt{2}$ . Since these values are unequal, we cannot achieve the second set of numbers from the first set.

Sum of squares of the three numbers in the triple is invariant. Then, the rest follows easily as the sum of squares of the numbers in the two triples is different!! So, not possible to get the second triple.