7. Given that $|a|, |b|$ are integers, and satisfy
$$
\begin{array}{l}
(|a|+|b|)(|a|+3|b|)=105, \\
(a+b) \cdot(a+3 b)=33 .
\end{array}
$$

Then the angle between $\boldsymbol{a}$ and $\boldsymbol{b}$ is ( ).
(A) $30^{\circ}$
(B) $60^{\circ}$
(C) $120^{\circ}$
(D) $150^{\circ}$
7. C.

Let $|\boldsymbol{a}|+|\boldsymbol{b}|=m$. Then $m(m+2|\boldsymbol{b}|)=105$.
Given $|b|<m \leqslant 10$, we have
$|b|=\frac{105}{2 m}-\frac{m}{2}<m \Rightarrow m \geqslant 6$.
Also, $105=3 \times 5 \times 7$, so $m=7$.
Thus, $|b|=\frac{105-49}{14}=4,|a|=7-4=3$.
Therefore, $(a+b) \cdot(a+3 b)$
$$
=|a|^{2}+3|b|^{2}+4|a||b| \cos \alpha=33 \text {, }
$$

where $\alpha$ is the angle between $\boldsymbol{a}$ and $\boldsymbol{b}$. Hence,
$$
\cos \alpha=\frac{33-|a|^{2}-3|b|^{2}}{4|a||b|}=-\frac{1}{2} .
$$

Given $0^{\circ} \leqslant \alpha<180^{\circ}$, then $\alpha=120^{\circ}$.