设 overrightarrow(a) = (2, -3, 2), overrightarrow(b) = (-4, 6, 4), 则 cos(overrightarrow(a), overrightarrow(b)) = ( )

根据题意，向量 $\vec{a} = (2, -3, 2)$，$\vec{b} = (-4, 6, 4)$。 首先，计算两向量的点积： \[ \vec{a} \cdot \vec{b} = 2 \times (-4) + (-3) \times 6 + 2 \times 4 = -8 - 18 + 8 = -18 \] 接着，计算两向量的模长： \[ |\vec{a}| = \sqrt{2^2 + (-3)^2 + 2^2} = \sqrt{4 + 9 + 4} = \sqrt{17} \] \[ |\vec{b}| = \sqrt{(-4)^2 + 6^2 + 4^2} = \sqrt{16 + 36 + 16} = \sqrt{68} = 2\sqrt{17} \] 利用余弦公式： \[ \cos(\vec{a}, \vec{b}) = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{-18}{\sqrt{17} \times 2\sqrt{17}} = \frac{-18}{34} = -\frac{9}{17} \] 因此，答案为： \[ \boxed{-\frac{9}{17}} \]