3. Given the function $f(x)=x^{2}-4 x+3$, the sets are
$$
\begin{array}{l}
M=\{(x, y) \mid f(x)+f(y) \leqslant 0\}, \\
N=\{(x, y) \mid f(x)-f(y) \geqslant 0\} .
\end{array}
$$

Then in the Cartesian coordinate system, the area of the region represented by the set $M \cap N$ is ( ).
(A) $\frac{\pi}{4}$
(B) $\frac{\pi}{2}$
(C) $\pi$
(D) $2 \pi$
3. C.

From the given information, we have
$$
\begin{array}{l}
M=\{(x, y) \mid f(x)+f(y) \leqslant 0\} \\
=\left\{(x, y) \mid(x-2)^{2}+(y-2)^{2} \leqslant 2\right\}, \\
N=\{(x, y) \mid f(x)-f(y) \geqslant 0\} \\
=\{(x, y) \mid(x-y)(x+y-4) \geqslant 0\} .
\end{array}
$$

Then, $M \cap N=\left\{\begin{array}{l}(x-2)^{2}+(y-2)^{2} \leqslant 2, \\ (x-y)(x+y-4) \geqslant 0 .\end{array}\right.$
The intersection part is shown in Figure 4, and its area is half of the circle's area, which is
$$
\frac{1}{2} \pi(\sqrt{2})^{2}=\pi \text {. }
$$