Let $0\le x,y,z,u,v,w\le n$ be integer numbers obeying
\begin{align*} x^2+y^2+z^2=&u^2+v^2+w^2\\ x+y+v=&u+w+z\\ x\neq& w \end{align*} (Please note that the second equality is $x+y+v=u+w+z$ NOT $x+y+z=u+v+w$. This has lead to some mistakes in some of the answers below)
How can the solutions to the above equations be characterized. One class of solutions is as follows
\begin{align*} x=&u\\ y=&w\\ z=&v \end{align*}
This system is well studied. You can find full description of solutions in "Introduction to the theory of numbers" by Leonard E. Dickson. (See Theorem 47).
If all the variables are between $1$ and $P$ then the number of solutions is $$\frac{18}{\pi^2}P^3\log P+O(P^3),$$ see "An asymptotic formula for the number of solutions of a system of equations" N.N. Rogovskaya - Diophantine approximations, Part II (Russian), 1986. She used more natural parametrization: if $a_1=x-u$, $a_2=y-v$, $a_3=z-w$, $b_1=x+u$, $b_2=y+v$, $b_3=z+w$, then the system is equivalent to $$a_1+a_2+a_3=0,\qquad a_1b_1+a_2b_2+a_3b_3=0.$$ All solutions of the equation $$kb_1+lb_2-(k+l)b_3=0$$ for $(k,l)=1$ are $$b_1=b_3-lt,\quad b_2=b_3+kt\quad(b_3,t\in\mathbb{Z}).$$
Your system can be also studied with circle method. And this is unique case (among Vinogradov's type systems) when trigonometric integal is known explicitly (from Rogovskaya's result). The circle method gives $$\frac{12}{\pi^2}\gamma P^3\log P+O(P^3),$$ where $$\gamma=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\left| \int_{0}^{1}e^{2\pi i(z_1x^2+z_2x)}dx\right|^6dz_1dz_2$$
UPD: This arguments are valid for symmetrical system of equations $$x^2+y^2+z^2=u^2+v^2+w^2,\qquad x+y+z=u+v+w.$$ The original system will take this form after changing variables $z\to -z$, $v\to -v$.
This may not give all the solutions, but it does give a 3-parameter family. Choose positive integers $a,b,c,d$ such that $$a+b=c+d,\quad a+c<b<a+d,\quad2a<b$$ Then $$b^2+(b-a-c)^2+(a+d-b)^2=c^2+d^2+(b-2a)^2$$ and $$b+(b-2a)+(a+d-b)=c+d+(b-a-c)$$
We need to solve the following system of two diophantine equations::
$$\quad x^2+y^2+z^2\ =\ u^2+v^2+w^2.$$ $$x+y+v\ =\ u+w+z$$
I will present some (not all) solutions. Naturally, there is more than one way to select only a special set, and I intend to present more than one (still far from all).
@Alexey has provided a complete Answer. This perhaps still has left some room for entertaining solutions. (Especially, that they may carry a combinatorial/geometric bonus).
FIRST METHOD Let's search for the pairs of triples $\ (x\ y\ z)\ \ (u\ v\ w),\ $ which are related to the Pythagorean triples as follows:
This would imply the first equality: $\ x^2+y^2+z^2\ =\ u^2+v^2+w^2.\ $ Then we only need to assure equality of perimeters of our rectangular triangles:
$$\ x+y+v\ =\ u+w+z$$
Now let's use the popular parametrization (it satisfies 1. and 2.):
The equality to be assurred (see the above centered formula) now spells:
$$ b\cdot(a+b)\ = s\cdot(r+s)$$
Thus for every two factorizations of an integer into positive factors:
$$ b\cdot c = s\cdot t\qquad(b<c<2\!\cdot\! b\quad and\quad s<t<2\!\cdot\! s) $$
and letting $\ a:=c\!-\!b\ \ \ r:=t\!-\!s,\ $ we obtain the respective two requested triples.
EXAMPLE (FIRST METHOD) Consider the following two factorizations of the same positive integer:
$$ 15\cdot 28\ =\ 20\cdot 21 $$
(We see that the larger factors on each side are smaller than their doubled partner: $\ 15<28<2\cdot 15\ $ and $\ 20<21<2\cdot 2\cdot 20$). Now:
$$a:=13\ \ b:=15\qquad r:=1\ \ s:=20$$ $$x:=56\ \ y:=390\ \ v:=394\qquad u:=399\ \ w:=40\ \ z:=401$$
This is a solution; visually:
SECOND METHOD (under construction)
OPTIONAL STUFF Now--for the sake of completeness--I will present certain solutions of a system of equations similar to (but different from) the one from the Question; this is a more symmetric system:
$$\quad x^2+y^2+z^2\ =\ u^2+v^2+w^2.$$ $$x+y+z\ =\ u+v+w$$
The idea is simple. Let's search for the pairs of triples $\ (x\ y\ z)\ \ (u\ v\ w),\ $ which are related to the Pythagorean triples as follows:
This would imply the first equality: $\ x^2+y^2+z^2\ =\ u^2+v^2+w^2.\ $ Then we only need to assure equality
$$\ x+y-u\ =\ v+w-z$$
Now let's use the popular parametrization (it satisfies 1. and 2.):
The equality to be assurred (see the above centered formula) now spells:
$$ a\cdot(b-a)\ = r\cdot (s-r)$$
Thus for every two factorizations of an integer into positive factors:
$$ a\cdot c = r\cdot t$$
and letting $\ b:=a\!+\!c\ \ s:=r\!+\!t,\ $ we obtain the respective two requested triples.
EXAMPLE The simplest of the kind (let's not worry here too much about the triples being relatively prime):
$$1\cdot 4\ =\ 2\cdot 2$$ $$a:=1\ \ b:=b':=5\ \ (and\ \ a':=4)\qquad r:=2\ \ s:=4$$
$$x=24\ \ \ y=10\ \ \ u=26\qquad(or\ \ x'\!=9\ \ \ y'\!=40\ \ \ u'\!=41)$$ $$z=20\ \ \ v=12\ \ w=16$$
Thus we get:
as well as
BONUS By considering--as above--multiple factorizations of any fixed (but easily factorable) integer, we obtain collections of equalities which have patterns of equalities of members of different triples (potentially of some combinatorial/geometric value).
For the system of equations:
$$\left\{\begin{aligned}&x^2+y^2+z^2=u^2+w^2+v^2\\&x+y+v=u+w+z\end{aligned}\right.$$
You can record solutions:
$$x=s^2+kt-ks+kq+ts-tq-qs$$
$$y=s^2+kt+ks-kq-ts+tq-qs$$
$$z=s^2+kt+ks-kq+ts-tq+2q^2-3qs$$
$$u=s^2+kt+ks-kq+ts-tq-qs$$
$$w=kt-s^2-ks+kq-ts+tq-2q^2+3qs$$
$$v=kt-s^2+ks-kq+ts-tq+qs$$
$s,q,k,t$ - integers asked us.
For not a lot of other systems of equations:
$$\left\{\begin{aligned}&x^2+y^2+z^2=u^2+w^2+v^2\\&x+y+z=u+w+v\end{aligned}\right.$$
Solutions have the form:
$$x=2s^2+(2q+2t+k)s+kt+qk+2qt$$
$$y=s^2+(q+k)s+qk+kt-t^2$$
$$z=s^2+(t+k)s+qk+kt-q^2$$
$$u=s^2+(q+t+k)s+qt+qk+kt$$
$$w=2s^2+(2q+2t+k)s+qt+qk+kt$$
$$v=s^2+ks+qk+kt-q^2-t^2$$
