12. Let $n$ be a positive integer, when $n>100$, the first two digits of the decimal part of $\sqrt{n^{2}+3 n+1}$ are
Rong Ge 49.
Analysis Because
$$
\frac{1}{\sqrt{n^{2}+3 n+1}-n-1}-2=\frac{\sqrt{n^{2}+3 n+1}+n+1}{n}-2=\sqrt{\frac{1}{n^{2}}+\frac{3}{n}+1}-1+\frac{1}{n}>0
$$

When $n>100$,
$$
\sqrt{\frac{1}{n^{2}}+\frac{3}{n}+1}-1+\frac{1}{n}<1+\frac{2}{n}-1+\frac{1}{n}=\frac{3}{n}<\frac{3}{101}<\frac{2}{49}
$$

Therefore, $0.49<\sqrt{n^{2}+3 n+1}-n-1<0.5$, so the first two digits of its decimal part are 49.