12. Let $n \in \mathbf{Z}_{+}$. When $n>100$, the first two decimal places of the fractional part of $\sqrt{n^{2}+3 n+1}$ are $\qquad$ .
12. 49.
$$
\begin{array}{l}
100 n+49<\sqrt{100^{2}\left(n^{2}+3 n+1\right)} \\
<100 n+150
\end{array}
$$