satisfying axioms to 'prove' vector space

hello there!

I need some clarity on the following please:

The question asked that it be shown that the following set is not a vector space:

'The set of all triples of real numbers with addition defined by

(*x, y, z*) + (*u, v, w*) = (*z+w, y+v, x+u*)'

I worked out that the following axiom holds:

(*k+m*)**u **= *k***u + ***k***u**

__however__, the answer works it out to *not* hold.

It goes as follows:

Set **u** = (*u1, u2, u3*), then

(*k+m*)**u**

**= **(*k+m*)(*u1, u2, u3*)

= ((*k+m*)*u1, *(*k+m*)*u2, *(*k+m*)*u3)*

*and *

*k***u **+ m*u*

*= *(k*u1, k**u2, ku3*) + (m*u1, mu2, mu3*)

= (k*u3 + mu3, k**u2 + mu2, ku1 *+ m*u1*)

since these are not equal, this axiom does not hold.'

Now, I don't understand why these two are not equal?

Can we not further simplify:

(k*u3 + mu3, k**u2 + mu2, ku1 *+ m*u1*)

= ((*k+m*)*u3, *(*k+m*)*u2, *(*k+m*)*u1) *?

That way, does it not equal = ((*k+m*)*u1, *(*k+m*)*u2, *(*k+m*)*u3)) *?

or is it not equal because the order of the triple is different?

I'm quite confused and any help would be much appreciated!

Have a great day all (Happy)