Question 1209500
Let the initial amounts of water in containers \( A \), \( B \), and \( C \) be \( a \), \( b \), and \( c \) liters, respectively. 

### Step 1: Transfer water from \( A \) to \( B \)
\[
\text{Amount poured from } A \text{ to } B = \frac{1}{5}a
\]
After this transfer:
\[
A = a - \frac{1}{5}a = \frac{4}{5}a, \quad B = b + \frac{1}{5}a
\]

### Step 2: Transfer water from \( B \) to \( C \)
\[
\text{Amount poured from } B \text{ to } C = \frac{1}{6}(b + \frac{1}{5}a)
\]
After this transfer:
\[
B = b + \frac{1}{5}a - \frac{1}{6}(b + \frac{1}{5}a) = b + \frac{1}{5}a - \frac{1}{6}b - \frac{1}{30}a
\]
\[
C = c + \frac{1}{6}(b + \frac{1}{5}a)
\]

### Step 3: Transfer water from \( C \) to \( A \)
\[
\text{Amount poured from } C \text{ to } A = \frac{1}{6}[c + \frac{1}{6}(b + \frac{1}{5}a)]
\]
After this transfer:
\[
C = c + \frac{1}{6}(b + \frac{1}{5}a) - \frac{1}{6}[c + \frac{1}{6}(b + \frac{1}{5}a)]
\]
\[
A = \frac{4}{5}a + \frac{1}{6}[c + \frac{1}{6}(b + \frac{1}{5}a)]
\]

### Step 4: Final state
At the end, all containers have the same amount of water: \( 8.5 \) liters.

#### Equations:
1. \( A = 8.5 \):
   \[
   \frac{4}{5}a + \frac{1}{6}[c + \frac{1}{6}(b + \frac{1}{5}a)] = 8.5
   \]
2. \( B = 8.5 \):
   \[
   b + \frac{1}{5}a - \frac{1}{6}(b + \frac{1}{5}a) = 8.5
   \]
3. \( C = 8.5 \):
   \[
   c + \frac{1}{6}(b + \frac{1}{5}a) - \frac{1}{6}[c + \frac{1}{6}(b + \frac{1}{5}a)] = 8.5
   \]

We will solve the system of equations to find \( b \), the initial amount of water in container \( B \).

The initial amount of water in container \( B \) was \( \mathbf{8.5 \, \text{liters}} \). It turns out that all three containers started with the same amount of water, \( 8.5 \) liters.