3. A.

From the given conditions,
$$
\begin{array}{l}
a^{2}+2+2 \sqrt{2} a=(b c+2)+(b+c) \sqrt{2} \\
\Rightarrow\left\{\begin{array}{l}
a^{2}+2=b c+2, \\
2 a=b+c
\end{array}\right. \\
\Rightarrow(b+c)^{2}=4 a^{2}=4 b c \\
\Rightarrow(b-c)^{2}=0 \Rightarrow b=c .
\end{array}
$$

Thus, $a=b=c$. Therefore, $\angle A=60^{\circ}$.
3. Given $\triangle A B C$ with the three interior angles $\angle A, \angle B, \angle C$ and the corresponding side lengths $a, b, c$, where $a, b, c$ are rational numbers, and they satisfy
$$
(a+\sqrt{2})^{2}=(b+\sqrt{2})(c+\sqrt{2}) \text {. }
$$

Then $\angle A=(\quad)$.
(A) $60^{\circ}$
(B) $75^{\circ}$
(C) $90^{\circ}$
(D) $120^{\circ}$