Let $ a = e^{ix} $, $ b = e^{iy} $, and $ c = e^{iz} $. Then \begin{align*} a + b + c &= e^{ix} + e^{iy} + e^{iz} \\ &= (\cos x + \cos y + \cos z) + i (\sin x + \sin y + \sin z) \\ &= 0.\end{align*}Also, \begin{align*} \frac{1}{a} + \frac{1}{b} + \frac{1}{c} &= \frac{1}{e^{ix}} + \frac{1}{e^{iy}} + \frac{1}{e^{iz}} \\ &= e^{-ix} + e^{-iy} + e^{-iz} \\ &= [\cos (-x) + \cos (-y) + \cos (-z)] + i [\sin (-x) + \sin (-y) + \sin (-z)] \\ &= (\cos x + \cos y + \cos z) - i (\sin x + \sin y + \sin z) \\ &= 0.\end{align*}Hence, \[abc \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) = ab + ac + bc = 0.\]Note that $ \cos (2x - y - z) $ is the real part of $ e^{i (2 \alpha - \beta - \gamma)}, $ and \begin{align*} e^{i (2 \alpha - \beta - \gamma)} &= \frac{a^2}{bc} \\ &= \frac{a \cdot a}{-ab - ac} \\ &= \frac{a (-b - c)}{-ab - ac} \\ &= 1.\end{align*}Therefore, $ \cos (2x - y - z) = \boxed{1} $.