There are some milk in three vessels.`1/3` part of the first mixture is poured into second vessel. `1/4` part of the second mixture is poured into third vessel. `1/10` . part of the third mixture is poured into first vessel. Now all three vessels contains 9 - 9 ltmilk, then find the initial quantity of milk in all three vessels.

Answer

To solve the problem step by step, let's denote the initial quantities of milk in the three vessels as follows:

- Let the initial quantity in Vessel 1 be \( X \) liters.
- Let the initial quantity in Vessel 2 be \( Y \) liters.
- Let the initial quantity in Vessel 3 be \( Z \) liters.

### Step 1: Understand the Transfers
1. **Transfer from Vessel 1 to Vessel 2**:
- \( \frac{1}{3} \) of \( X \) is poured into Vessel 2.
- Remaining in Vessel 1: \( X - \frac{1}{3}X = \frac{2}{3}X \).
- New quantity in Vessel 2: \( Y + \frac{1}{3}X \).

2. **Transfer from Vessel 2 to Vessel 3**:
- \( \frac{1}{4} \) of the new quantity in Vessel 2 is poured into Vessel 3.
- New quantity in Vessel 2: \( Y + \frac{1}{3}X - \frac{1}{4}(Y + \frac{1}{3}X) \).
- Remaining in Vessel 2 after transfer:
\[ Y + \frac{1}{3}X - \frac{1}{4}(Y + \frac{1}{3}X) = \frac{3}{4}(Y + \frac{1}{3}X). \]
- New quantity in Vessel 3:
\[ Z + \frac{1}{4}(Y + \frac{1}{3}X). \]

3. **Transfer from Vessel 3 to Vessel 1**:
- \( \frac{1}{10} \) of the new quantity in Vessel 3 is poured back into Vessel 1.
- Remaining in Vessel 3:
\[ Z + \frac{1}{4}(Y + \frac{1}{3}X) - \frac{1}{10}(Z + \frac{1}{4}(Y + \frac{1}{3}X). \]
- New quantity in Vessel 1:
\[ \frac{2}{3}X + \frac{1}{10}(Z + \frac{1}{4}(Y + \frac{1}{3}X)). \]

### Step 2: Set Up the Equations
After all transfers, we know that each vessel contains 9 liters of milk. Therefore, we can set up the following equations:

1. For Vessel 1:
\[ \frac{2}{3}X + \frac{1}{10}(Z + \frac{1}{4}(Y + \frac{1}{3}X)) = 9. \]

2. For Vessel 2:
\[ \frac{3}{4}(Y + \frac{1}{3}X) = 9. \]

3. For Vessel 3:
\[ Z + \frac{1}{4}(Y + \frac{1}{3}X) - \frac{1}{10}(Z + \frac{1}{4}(Y + \frac{1}{3}X)) = 9. \]

### Step 3: Solve the Equations
1. **From Vessel 2**:
\[ \frac{3}{4}(Y + \frac{1}{3}X) = 9 \implies Y + \frac{1}{3}X = 12 \implies Y = 12 - \frac{1}{3}X. \]

2. **Substituting \( Y \) into Vessel 1**:
\[ \frac{2}{3}X + \frac{1}{10}\left(Z + \frac{1}{4}(12 - \frac{1}{3}X + \frac{1}{3}X)\right) = 9. \]
Simplifying gives:
\[ \frac{2}{3}X + \frac{1}{10}(Z + 3) = 9. \]

3. **From Vessel 3**:
\[ Z + \frac{1}{4}(12 - \frac{1}{3}X) - \frac{1}{10}(Z + 3) = 9. \]
Simplifying gives:
\[ Z + 3 - \frac{1}{10}Z - \frac{3}{10} = 9 \implies \frac{9}{10}Z + \frac{27}{10} = 9. \]
Solving for \( Z \):
\[ \frac{9}{10}Z = 9 - \frac{27}{10} \implies \frac{9}{10}Z = \frac{63}{10} \implies Z = 7. \]

4. **Substituting \( Z \) back into the equations**:
- From \( Y \):
\[ Y = 12 - \frac{1}{3}X. \]
- From Vessel 1:
\[ \frac{2}{3}X + \frac{1}{10}(7 + 3) = 9 \implies \frac{2}{3}X + 1 = 9 \implies \frac{2}{3}X = 8 \implies X = 12. \]
- Finally, substituting \( X \) back to find \( Y \):
\[ Y = 12 - \frac{1}{3}(12) = 12 - 4 = 8. \]

### Final Results
- Initial quantity in Vessel 1: \( X = 12 \) liters.
- Initial quantity in Vessel 2: \( Y = 8 \) liters.
- Initial quantity in Vessel 3: \( Z = 7 \) liters.

### Summary
The initial quantities of milk in the three vessels are:
- Vessel 1: 12 liters
- Vessel 2: 8 liters
- Vessel 3: 7 liters