15. Given $3 \vec{a}-2 \vec{b}=\{-2,0,4\}, \vec{c}=\{-2,1,2\}, \vec{a} \cdot \vec{c}=2,|\vec{b}|=4$. Find the angle $\theta$ between $\vec{b}$ and $\vec{c}$.
15. From the problem, we have $(3 \vec{a}-2 \vec{b}) \cdot \vec{c}=12$, and $(3 \vec{a}-2 \vec{b}) \cdot \vec{c}=3 \vec{a} \cdot \vec{c}-2 \vec{b} \cdot \vec{c}=6-2 \vec{b} \cdot \vec{c}$, thus $\vec{b} \cdot \vec{c}=-3, \vec{b} \cdot \vec{c}=|\vec{b}||\vec{c}| \cos \theta=12 \cos \theta, \cos \theta=-\frac{1}{4} \theta=\pi-\arccos \frac{1}{4}$.