3. There are three vessels with capacities of 6, 3, and 7 liters, respectively. The first vessel contains 4 liters, and the third vessel contains 6 liters of milk. Using these three vessels, it is necessary to divide the milk equally between two vessels.
3. First, let's determine the total amount of milk:

$$
4+6=10 \text { (L). }
$$

Next, we need to find out how many liters should be in one container:

$$
\text { 10:2 = } 5 \text { (L). }
$$

Therefore, at the last pouring, the possible scenarios are $5-0-5, 5-5-0, 0-5-5$. The last two scenarios are impossible because 5 liters cannot be poured into a 3-liter container. Therefore, at the last pouring, we will have 5-0-5. The solution can be represented as a sequence of pourings:

| 6L | 3L | 7L |
| :---: | :---: | :---: |
| 4 | 0 | 6 |
| 1 | 3 | 6 |
| 1 | 2 | 7 |
| 6 | 2 | 2 |
| 5 | 3 | 2 |
| 5 | 0 | 5 |