Question:

Step 1: From the given condition $\vec{a} \times \vec{b} + \vec{c} = 0$, we have $\vec{c} = -\vec{a} \times \vec{b} = \vec{b} \times \vec{a}$. Step 2: Since $\vec{c}$ is normal (perpendicular) to $\vec{b}$, we have $\vec{c} \cdot \vec{b} = 0$. This means $(\vec{b} \times \vec{a}) \cdot \vec{b} = 0$, which is always true by the cross product property. Step 3: For $\vec{c} \neq \vec{0}$ to be perpendicular to $\vec{b}$, vectors $\vec{a}$ and $\vec{b}$ cannot be parallel. However, the constraint $\vec{a} \cdot \vec{b} = 3$ (non-zero) combined with $\vec{c} \perp \vec{b}$ forces $\vec{b}$ to be parallel to $\vec{a}$. Step 4: If $\vec{b} = \lambda\vec{a}$ for some scalar $\lambda > 0$, then $\vec{a} \cdot \vec{b} = \vec{a} \cdot (\lambda\vec{a}) = \lambda|\vec{a}|^2 = 3$, and $\vec{b} \times \vec{a} = \lambda\vec{a} \times \vec{a} = \vec{0}$, making $\vec{c} = \vec{0}$ which satisfies all conditions. Step 5: Therefore, $\vec{b}$ is parallel to $\vec{a}$ with the specific relationship determined by $\vec{a} \cdot \vec{b} = 3$. ∴ Answer: A