3. Given the sequence $\left\{a_{n}\right\}$ satisfies $3 a_{n+1}+a_{n}=4(n \geqslant 1)$, and $a_{1}=9$, the sum of the first $n$ terms is $S_{n}$. Then the smallest integer $n$ that satisfies the inequality $\left|S_{n}-n-6\right|<\frac{1}{125}$ is
A. 5
B. 6
C. 7
D. 8
$$
\begin{array}{l}
a_{n+1}=-\frac{1}{3} a_{n}+\frac{4}{3} \Rightarrow a_{n+1}-1=-\frac{1}{3}\left(a_{n}-1\right) \Rightarrow a_{n}-1=\left(a_{1}-1\right)\left(-\frac{1}{3}\right)^{n-1} \\
\quad=8\left(-\frac{1}{3}\right)^{n-1} \Rightarrow a_{n}=1+8\left(-\frac{1}{3}\right)^{n-1} \Rightarrow S_{n}=n+\frac{8\left[1-\left(-\frac{1}{3}\right)^{n}\right]}{1+\frac{1}{3}} \\
\quad=n+6-6\left(-\frac{1}{3}\right)^{n} \Rightarrow\left|S_{n}-n-6\right|=\frac{6}{3^{n}} .
\end{array}
$$

Thus, $\frac{6}{3^{n}} < 750 \Rightarrow n \geqslant 7$, so the answer is $C$.