3. Given that $a$ and $b$ are constants, the solution set of the inequality $a b(x-1)>a x+b$ with respect to $x$ is $x<2$. Then the value of the algebraic expression
$$
|a-1|+|b-2|-|a+b-3|
$$

is ( ).
(A) 0
(B) $2 a-2$
(C) $2-2 a$
(D) $2 b-6$
3. A.

From the given, we know that $ab = 2a + b$.
Clearly, $b \neq 2$, and $a = \frac{b}{b-2}$.
Thus, $a-1 = \frac{2}{b-2}$.
Then, $a-1$ and $b-2$ have the same sign.
Therefore, $|a-1| + |b-2| - |a+b-3| = 0$.