3. Given that $x$ is a real number, and
$$
|3 x-1|+|4 x-1|+\cdots+|17 x-1|
$$

is a definite constant. Then this constant is ( ).
(A) 5
(B) 10
(C) 15
(D) 75
3.A.

Since the value of the original expression is a definite constant, removing the absolute value symbol should eliminate $x$.
$$
\begin{array}{l}
\text { and } 3+4+\cdots+12=13+14+\cdots+17 \text {, therefore, } \\
12 x-1 \leqslant 0 \text {, and } 13 x-1 \geqslant 0 \text {. } \\
\text { Solving, we get } \frac{1}{13} \leqslant x \leqslant \frac{1}{12} \text {. } \\
\text { Hence, the original expression }=(1-3 x)+(1-4 x)+\cdots+(1-12 x)+ \\
(13 x-1)+(14 x-1)+\cdots+(17 x-1) \\
=5 \text {. } \\
\end{array}
$$