3. In the Cartesian coordinate system, points whose both coordinates are rational numbers are called rational points. If $a$ is an irrational number, then among all lines passing through the point $(a, 0)$,
(A) there are infinitely many lines, each of which has at least two rational points:
(B) there are $n(2 \leqslant n<.+\infty)$ lines, each of which has at least two rational points;
(C) there is exactly one line that passes through at least two rational points;
(D) each line passes through at most one rational point.
Solution: The $x$-axis passes through the point $(a, 0)$, and the $x$-axis contains at least two rational points, so option (D) should be excluded.

Assume there is another line with a slope of $k (k \neq 0)$ that contains at least two rational points, let's say $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, then $x_{1}, x_{2}, y_{1}, y_{2} \in Q$, and $y_{1} \neq 0, x_{1} \neq x_{2}$. On one hand, $k=\frac{y_{1}-a}{x_{1}-a}$ $\in \bar{Q}$, and on the other hand, $k=\frac{y_{1}-y_{2}}{x_{1}-x_{2}} \in Q$, leading to a contradiction. Therefore, $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ cannot both be rational points.

When the slope of the line does not exist, the line $x=a$ does not contain any rational points. However, when $k=0$, the line $y=0$ (the $x$-axis) passing through the point $(a, 0)$ contains infinitely many rational points.
Thus, the correct choice is $(C)$.