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Let $x+y+z=i$

Then $(x+y)i=30 \\ (y+z)i=18 \\ (z+x)i=2$

Adding them we get, $2(x+y+z)i=50 \\ i^{2}=25 \\ i=\pm{5}$

Keeping $i$ 's value in the first equation of $x+y$ , we get $x+y=\pm{6}$

$x+y+z=\pm{5} \\ x+y=\pm{6} \\ z=\pm{5}-(\pm{6})$

We see $z=\pm{11},\pm{1}$ adding all we get answer to be $\boxed{0}$

Tricky solution:

Note that if $(x_0, y_0, z_0)$ satisfy the system of equations, then so does $(-x_0, -y_0, -z_0)$ .

So the sum of all possible values of $z$ must be $\boxed{0}$ .

let $x+y+z=a$ then, $\begin{cases} a(a-z)=30\rightarrow a^2-az=30\\ a(a-x)=18\rightarrow a^2-ax=18\\ a(a-y)=2\rightarrow a^2-az=2 \end{cases}$ adding them: $3a^2-a(x+y+z)=3a^2-a*a=2a^2=50$ $a=\pm 5$ insert this into 1st equation: $(\pm 5)^2-\pm 5z=30$ $\pm 5z=-5\rightarrow z=-\pm 1$ $1+-1=\boxed{0}$