114 Let the function $y=f(x)$ satisfy for all real numbers $x$
$$
f(3+x)=f(3-x)
$$

and the equation $f(x)=0$ has exactly 6 distinct real roots, then the sum of these 6 roots is
A. 18
B. 12
C. 9
D. 0
114 A. If $3+\alpha$ is a root of $f(x)=0$, then by the given condition we know
$$
f(3-\alpha)=f(3+\alpha)=0,
$$

which means $3-\alpha$ is also a root of $f(x)=0$. Therefore, we can assume these 6 roots are
$$
3 \pm \alpha_{1}, 3 \pm \alpha_{2}, 3 \pm \alpha_{3} .
$$

Thus, their sum is 18. Hence, the answer is $\mathrm{A}$.