3. If the set $M=\left\{x \left\lvert\, \frac{|3-x|}{|5-x|} \leqslant \frac{1}{2}\right.\right\}$ and the set $N=\left\{x \mid x^{2}-2 x+c \leqslant 0\right\}$ satisfy $M \cap N=M$, then the range of the real number $c$ is ( ).
(A) $c \leqslant-\frac{44}{9}$
(B) $c \leqslant-\frac{55}{9}$
(C) $c \leqslant-\frac{66}{9}$
(D) $c \leqslant-\frac{77}{9}$
3. B.
$$
\begin{array}{l}
\text { Given } M=\left\{x \neq 5 \mid 4(3-x)^{2} \leqslant(5-x)^{2}\right\} \\
=\left\{x \neq 5 \mid 3 x^{2}-14 x+11 \leqslant 0\right\} \\
=\left\{x \left\lvert\, 1 \leqslant x \leqslant \frac{11}{3}\right.\right\}, \\
N=\{x \mid 1-\sqrt{1-c} \leqslant x \leqslant 1+\sqrt{1-c}\},
\end{array}
$$

and $1-\sqrt{1-c} \leqslant 1$,
we get $1+\sqrt{1-c} \geqslant \frac{11}{3} \Rightarrow c \leqslant-\frac{55}{9}$.