(7) Let the set $S=\{1,2, \cdots, 15\}, A=\left\{a_{1}, a_{2}, a_{3}\right\}$ be a subset of $S$, and $(a_{1}, a_{2}, a_{3})$ satisfy:
$$
1 \leqslant a_{1}<a_{2}<a_{3} \leqslant 15, a_{3}-a_{2} \leqslant 6 .
$$

Then the number of subsets $A$ that satisfy the condition is $\qquad$
(7) When $2 \leqslant a_{2} \leqslant 9$, $\left(a_{1}, a_{2}\right)$ has $\mathrm{C}_{9}^{2}$ ways to choose, and $a_{3}$ has 6 ways to choose, so $\left(a_{1}, a_{2}, a_{3}\right)$ has a total of $6 \times \mathrm{C}_{9}^{2}=216$ (ways) to choose; When $10 \leqslant a_{2} \leqslant 14$, once $a_{2}$ is determined, $a_{1}$ has $a_{2}-1$ ways to choose, and $a_{3}$ has $15-a_{2}$ ways to choose, so the number of ways to choose $\left(a_{1}, a_{2}, a_{3}\right)$ is
$$
\begin{aligned}
& \sum_{a_{2}=10}^{14}\left(a_{2}-1\right)\left(15-a_{2}\right) \\
= & 9 \times 5+10 \times 4+11 \times 3+12 \times 2+13 \times 1 \\
= & 155 \text { (ways). }
\end{aligned}
$$

In summary, the number of subsets $A$ that satisfy the conditions is $216+155=371$ (subsets).