2. Let the solution set of $x^{2}-5 x+6 \leqslant 0$ be $P$, and the solution set of $x^{2}-2 a x+a^{2}-1 \leqslant 0$ be $Q$. Given that $P \subset Q$. Then the range of real number $a$ is $\qquad$
$2.2 \leqslant a \leqslant 3$.
From the problem, we know
$$
\begin{array}{l}
P=\{x \mid 2 \leqslant x \leqslant 3\}, \\
Q=\{x \mid a-1 \leqslant x \leqslant a+1\} .
\end{array}
$$

Since $P \subset Q$, therefore, $a-1 \leqslant 2$, and $3 \leqslant a+1$. Solving this, we get $2 \leqslant a \leqslant 3$.