When you ran out of ideas... again

First, by adding equations 1, 3, 5, we have

$(a+b+c)=(b+c+a)z+(c+a+b)x$

so that we know $z=1-x$

From this, we solve for $x$ in equations 1, 3, and then solve for $c$ from the two

$c = \dfrac{1}{2}\left( a + b - \sqrt{3} \sqrt{ -a^2 +2ab -b^2 }\right)$

so that $c$ is real only if $a=b$ . This is the crux step. Immediately from this, we see that $a=b=c=y$

From this, we then solve for $x$ and $a$ from equations 2, 4, and find that $x = a = \dfrac{1}{2}$ . Hence

$(a+b+c)(x+y+z) = \dfrac{9}{4}$

Another possible solution would be

$(a, b, c, x, y, z)=(-1, -1, -1, \frac{1}{2}, -1, \frac{1}{2})$ which would have yielded $0$