13.192. There are three vessels containing unequal amounts of liquid. To equalize these amounts, three pourings were made. First, $1 / 3$ of the liquid was poured from the first vessel into the second, then $1 / 4$ of the liquid that ended up in the second vessel was poured into the third, and finally, $1 / 10$ of the liquid that ended up in the third vessel was poured into the first. After this, each vessel contained 9 liters of liquid. How much liquid was initially in each vessel
Solution.

Let's form a table:

| Cо-  container | Initial  amount  of  liquid | After  the  first  transfer | After the second  transfer | After the third  transfer |
| :---: | :---: | :---: | :---: | :---: |
| 1 | $x$ l | $\frac{2}{3} x$ l | $\frac{2}{3} x \pi$ | $\frac{2}{3} x+\frac{1}{10}\left(z+\frac{1}{4}\left(y+\frac{1}{3}\right)\right) \pi$ |
| 2 | $y \pi$ | $y+\frac{1}{3} x \pi$ | $\frac{3}{4}\left(y+\frac{1}{3} x\right) \pi$ | $\frac{3}{4}\left(y+\frac{1}{3} x\right) \pi$ |
| 3 | $z \pi$ | $z \pi$ | $z+\frac{1}{4}\left(y+\frac{1}{3} x\right) \pi$ | $\frac{9}{10}\left(z+\frac{1}{4}\left(y+\frac{1}{3} x\right)\right) \pi$ |

Solving the system $\left\{\begin{array}{l}\frac{2}{3} x+\frac{1}{10}\left(z+\frac{1}{4}\left(y+\frac{1}{3} x\right)\right)=9, \\ \frac{3}{4}\left(y+\frac{1}{3} x\right)=9, \\ \frac{9}{10}\left(z+\frac{1}{4}\left(y+\frac{1}{3} x\right)\right)=9,\end{array} \quad\right.$ we find $x=12$ l;

$y=8 \pi ; z=7$ l.

Answer: 12, 8, and 7 l.