12. (10 points) In a math competition, each team can only score 0 points, 3 points, or 5 points per question. At the end of the competition, the total score of three teams is 32 points. If any team's total score can reach 32 points, how many different combinations of total scores are there for these three teams?
【Analysis】Let the number of questions worth 3 points be $x$, and the number of questions worth 5 points be $y$. Then we can get the indeterminate equation: $3 x+5 y=32$, $y=\frac{32-3 x}{5}$, so $x=9, y=1$ or $x=4, y=4$; then we can discuss the two cases separately.

【Solution】Solution: Let the number of questions worth 3 points be $x$, and the number of questions worth 5 points be $y$, then: $3 x+5 y=32$, so there are only two cases: $x=9, y=1$ or $x=4, y=4$:
(1) When $x=9, y=1$, it is equivalent to the three teams dividing 9 questions worth 3 points and 1 question worth 5 points, the number of ways the three teams can divide the 5 points is 3,

When $0 \leqslant a \leqslant 9$ (where $a$ represents the number of scores), if a team gets $a$ 3 points, then the number of ways the other two teams can divide $(9-a)$ 3 points is (10-a),

So for 9 questions worth 3 points, there are: $10+9+8+\cdots+1=55$ (ways),
The total number of ways to divide 9 questions worth 3 points and 1 question worth 5 points is: $55 \times 3=165$ (ways).
(2) When $x=4, y=4$, it is equivalent to the three teams dividing 4 questions worth 3 points and 4 questions worth 5 points,
When $0 \leqslant a \leqslant 4$, if a team gets $a$ 3 points, then the number of ways the other two teams can divide $(4-a)$ 3 points is (5a),

So for 4 questions worth 3 points, there are: $5+4+3+2+1=15$ (ways);
Similarly, when the three teams divide 4 questions worth 5 points, there are also 15 ways, but when a team gets no less than 3 questions worth 5 points, the 3 questions worth 5 points are equivalent to 5 questions worth 3 points in (1), which has already been considered in (1), and the number of ways the three teams can divide 4 questions worth 5 points, with one team getting no less than 3 questions, is 9, so the number of ways the three teams can divide 4 questions worth 3 points and 4 questions worth 5 points is: $15 \times(15-9)=90$ (ways); combining (1) and (2), the total number of different ways the three teams can score is: $165+90=255$ (ways).

Answer: The total number of different scoring situations for the three teams is 255.