13.402 The capacities of three vessels $A$, B, C, each of which has the shape of a cube, are in the ratio $1: 8: 27$, and the volumes of water poured into them are in the ratio 1:2:3. After transferring part of the water from vessel A to vessel B and from vessel B to vessel C, the water level in all three vessels became the same. Then, $128 \frac{4}{7}$ liters of water were transferred from vessel C to vessel B, and after that, from vessel B to vessel A, so that the water depth in vessel A became twice that in vessel B. As a result, it turned out that there was now $10 \theta$ liters less water in vessel A than there was initially. How much water was there initially in each vessel?
Solution. Let $x$ be the initial volume of water in vessel $\mathrm{A}$, and $a$ be the linear size of vessel A. According to the problem, $2x$ and $3x$ are the volumes of water in B and C, and $2a$ and $3a$ are the linear sizes of B and C. Since after the first transfer, the water level in all vessels became the same, the volumes of water were distributed in the ratio $a^2: 4a^2: 9a^2 = 1: 4: 9$. Therefore, the volume of water in A became $\frac{6x}{1+4+9}$, and the volume of water in B became $\frac{6x}{14} \cdot 4$. In total, in A and B, there were $\frac{15}{7} x$ liters. After transferring from C, this amount increased by $128 \frac{4}{7}$ liters. Finally, after the last transfer, the volumes of water in A and B (according to the problem) were distributed in the ratio $2a^2: 4a^2 = 1: 2$, i.e., in A there were $\frac{1}{3}\left(\frac{15}{7} x + \frac{900}{7}\right)$ liters.

From this, $\frac{1}{3}\left(\frac{15}{7} x + \frac{900}{7}\right) + 100 = x \Rightarrow x = 500$.

Answer: 500 liters, 1000 liters, 1500 liters.