(3) Given $f(x)=\left\{\begin{array}{ll}(3-a) x-a, & x<1, \\ \log _{a} x, & x \geqslant 1\end{array}\right.$ is monotonically increasing on $(-\infty,+\infty)$, then the range of values for $a$ is ( ).
(A) $(0,3)$
(B) $\left(1, \frac{3}{2}\right]$
(C) $\left[\frac{3}{2}, 3\right)$
(D) $(1,3)$
(3) From the problem, we have $\left\{\begin{array}{l}a>1, \\ 3-a>0, \\ \log _{a} 1 \geqslant(3-a) \cdot 1-a,\end{array}\right.$ solving this yields $\frac{3}{2} \leqslant a<3$. Therefore, the correct choice is C.