12. (5 points) Three people, A, B, and C, go fishing together. They put the fish they caught in a basket and lie down to rest, but they all fall asleep. A wakes up first, divides the fish in the basket into 3 equal parts, and finds 1 extra fish, which he throws back into the river. He then takes one part and goes home. B wakes up next, divides the remaining fish in the basket into 3 equal parts, and finds 1 extra fish, which he also throws back into the river. He then takes one part and goes home. C wakes up last and divides the remaining fish in the basket into 3 equal parts, and there is also 1 extra fish. These three people caught at least $\qquad$ fish.
【Solution】Solution: Let丙take away $a$ fish, 乙take away $b$ fish, and 甲take away $c$ fish, then:
When $a=1$, then $b=2$,
$3 b+1=3 \times 2+1=7 ; 7$ cannot be divided by 2, so $a \neq 1$;
When $a=2$, $3 a+1=3 \times 2+1=7 ; 7$ cannot be divided by 2, so $a \neq 2$;
When $a=3$,
$$
\begin{array}{l}
b=(3 a+1) \div 2, \\
=(3 \times 3+1) \div 2, \\
=10 \div 2, \\
=5 \text { (fish); } \\
c=(3 b+1) \div 2, \\
=(3 \times 5+1) \div 2, \\
=16 \div 2, \\
=8 \text { (fish); }
\end{array}
$$

So originally there were:
$$
8 \times 3+1=25 \text { (fish); }
$$

Answer: These three people caught at least 25 fish.