Let the first pile have Bai Zi X 1, then the sunspot is 80-X 1.

Let the second pile have Bai Zi X2, then the sunspot is 80-x2.

According to the known conditions, the number of sunspots in the third pile is 80*( 1-3/5)=32.

And because there are as many sunspots in the first pile of Bai Zi as in the second pile, if there is X 1=80-X2, X 1+X2=80.

Then the number of sunspots in the * * * three piles is 80-x1+80-x2+32 =192-x1+x2 =12.