\(\vec{a} \times \vec{b}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\), so it is also perpendicular to any combination of them, like \(\vec{a} + \vec{b}\) or \(\vec{a} - \vec{b}\). So options (A), (B) and (D) are always perpendicular.

For (C): \((\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 - |\vec{b}|^2\). This is zero only if \(|\vec{a}| = |\vec{b}|\). In general it is not zero, so these two vectors are not perpendicular.

Answer: (C)