1. **State the problem:** Given vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ with $\mathbf{a} \neq \mathbf{0}$, and the equation $\mathbf{a} \times 3\mathbf{b} = 2 \mathbf{a} \times \mathbf{c}$, show that $3\mathbf{b} - 2\mathbf{c} = k\mathbf{a}$ for some scalar constant $k$. 2. **Recall the vector cross product properties:** - The cross product $\mathbf{u} \times \mathbf{v}$ is perpendicular to both $\mathbf{u}$ and $\mathbf{v}$. - If $\mathbf{a} \times \mathbf{d} = \mathbf{0}$ and $\mathbf{a} \neq \mathbf{0}$, then $\mathbf{d}$ must be parallel to $\mathbf{a}$, i.e., $\mathbf{d} = k\mathbf{a}$ for some scalar $k$. 3. **Rewrite the given equation:** $$\mathbf{a} \times 3\mathbf{b} = 2 \mathbf{a} \times \mathbf{c}$$ 4. **Bring all terms to one side:** $$\mathbf{a} \times 3\mathbf{b} - 2 \mathbf{a} \times \mathbf{c} = \mathbf{0}$$ 5. **Use distributive property of cross product:** $$\mathbf{a} \times (3\mathbf{b} - 2\mathbf{c}) = \mathbf{0}$$ 6. **Since $\mathbf{a} \neq \mathbf{0}$ and the cross product is zero, the vector inside the parentheses must be parallel to $\mathbf{a}$:** $$3\mathbf{b} - 2\mathbf{c} = k\mathbf{a}$$ 7. **Conclusion:** We have shown that $3\mathbf{b} - 2\mathbf{c}$ is a scalar multiple of $\mathbf{a}$, as required. **Final answer:** $$3\mathbf{b} - 2\mathbf{c} = k\mathbf{a}$$