From \(\vec{a} \times \vec{c} + \vec{b} = \vec{0}\), we get \(\vec{a} \times \vec{c} = -\vec{b}\). Taking dot product with \(\vec{a}\): \(\vec{a} \cdot (\vec{a} \times \vec{c}) = -\vec{a} \cdot \vec{b}\). Since \(\vec{a} \cdot (\vec{a} \times \vec{c}) = 0\), we have \(\vec{a} \cdot \vec{b} = 0\), which checks: \((1)(-1)+(-1)(1)+(0)(1) = 0\). Nope: \(\vec{a} = \hat{i}-\hat{j}\) (only 2D) and \(\vec{b} = \hat{i}+\hat{j}+\hat{k}\), so \(\vec{a} \cdot \vec{b} = 1-1 = 0\). Good. Taking magnitude: \(|\vec{a} \times \vec{c}| = |\vec{b}| = \sqrt{3}\). Also \(|\vec{a} \times \vec{c}|^2 = |\vec{a}|^2|\vec{c}|^2 - (\vec{a} \cdot \vec{c})^2\), so \(3 = 2|\vec{c}|^2 - 16\), giving \(|\vec{c}|^2 = \frac{19}{2}\).