3. Given the sequence $\left\{a_{n}\right\}$ satisfies $3 a_{n+1}+a_{n}=4(n \geqslant 1)$, and $a_{1}=9$, with the sum of its first $n$ terms being $S_{n}$. Then the smallest integer $n$ that satisfies the inequality $\left|S_{n}-n-6\right|<\frac{1}{125}$ is $\qquad$.
3. $n=7$ Detailed Explanation: From the recurrence relation, we get: $3\left(a_{n+1}-1\right)=-\left(a_{n}-1\right)$, thus $\left\{a_{n}-1\right\}$ is a geometric sequence with the first term 8 and common ratio $-\frac{1}{3}$. Therefore, $S_{n}-n=\left(a_{1}-1\right)+\left(a_{2}-1\right)+\cdots+\left(a_{n}-1\right)=\frac{8\left[1-\left(-\frac{1}{3}\right)^{n}\right]}{1+\frac{1}{3}}=6-6 \times\left(-\frac{1}{3}\right)^{n}$, $\therefore\left|S_{n}-n-6\right|=6 \times\left(\frac{1}{3}\right)^{n}250, \therefore$ the smallest integer $n$ that satisfies the condition is $n=7$.