12. Given the line $a x+b y+c=0$ where $a, b, c$ are 3 different elements taken from the set $\{-3,-2$, $-1,0,1,2,3\}$, and the inclination angle of the line is acute. Then the number of such lines is $\qquad$
12. 43 Since the inclination angle of the line is acute, $\tan \theta=-\frac{a}{b}>0$, so $a$ and $b$ have opposite signs.

Note that when $a_{1}: b_{1}: c_{1}=a: b: c$, $a_{1} x+b_{1} y+c_{1}=0$ and $a x+b y+c=0$ represent the same line.
Assume without loss of generality that $a>0, b<0$.
(1) $c=0$, $a$ and $b$ each have 3 possible values, but among them, $x-y=0, 2x-2y=0, 3x-3y=0$ are the same line, so there are $3 \times 3-2=7$ such lines.
(2) $c \neq 0$, then $a$ and $b$ still each have 3 possible values. When $a$ and $b$ are determined, since $c, a, b$ are distinct elements, $c$ has 4 possible values, and there will be no different sets of values representing the same line. Therefore, there are $3 \times 3 \times 4=36$ such lines.
Thus, the total number of lines that meet the conditions is $7+36=43$.