3. (7th "Hope Cup" Invitational Training Question) Given $a^{2}+b^{2}+c^{2}=4, x^{2}+y^{2}+z^{2}=9$, what is the range of values for $a x + b y + c z$?
3. Let $\vec{a}=(a, b, c), \vec{b}=(x, y, z)$, then
$$
\begin{array}{l}
4=a^{2}+b^{2}+c^{2}=|\vec{a}|^{2} \geqslant \frac{(a \cdot \vec{b})^{2}}{|\vec{b}|^{2}}=\frac{(a x+b y+c z)^{2}}{x^{2}+y^{2}+z^{2}}=\frac{(a x+b y+c z)^{2}}{9}, \\
(a x+b y+c z)^{2} \leqslant 36,
\end{array}
$$
The range of $a x+b y+c z$ is
$$
-6 \leqslant a x+b y+c z \leqslant 6 .
$$