Given:
Ratio of milk and water in 1st mixture = 2 : 3
Ratio of milk and water in 2nd mixture = 3 : 1
Weight of third mixture = 7 litres
Percentage of milk in third mixture = 70%

To find:
Weight of 2nd mixture required to make the third mixture

Solution:
Let's assume that x litres of the 2nd mixture is taken to make the third mixture.

From the given information, we know that the weight of the third mixture is 7 litres. Let's assume that the density of the mixture is 1 g/ml. Then, the weight of the mixture will be equal to its volume.

Now, we need to find the amount of milk in the third mixture. We know that the percentage of milk in the third mixture is 70%. Therefore, the amount of milk in the third mixture will be:

Amount of milk in third mixture = 70% of 7 litres = 4.9 litres

Let's assume that the ratio of milk and water in the third mixture is a : b. Then, we can write:

a/b = 70/30 (as the percentage of milk is 70%)

We can simplify this ratio as follows:

a/b = 7/3

This means that for every 7 units of milk in the third mixture, there are 3 units of water.

Now, let's find the amount of milk and water in the first mixture and the second mixture.

Amount of milk in the first mixture = 2/5 * weight of first mixture
Amount of water in the first mixture = 3/5 * weight of first mixture

Amount of milk in the second mixture = 3/4 * weight of second mixture
Amount of water in the second mixture = 1/4 * weight of second mixture

Now, let's assume that y litres of the first mixture is taken to make the third mixture.

Then, the amount of milk in the third mixture will be:

Amount of milk in third mixture = Amount of milk in first mixture + Amount of milk in second mixture

Using the ratios, we can write:

2/5 * y + 3/4 * x = 7/10 * 7

Simplifying this equation, we get:

8y + 15x = 147

Similarly, the amount of water in the third mixture will be:

Amount of water in third mixture = Amount of water in first mixture + Amount of water in second mixture

Using the ratios, we can write:

3/5 * y + 1/4 * x = 3/10 * 7

Simplifying this equation, we get:

12y + 5x = 21

We now have two equations with two variables. Solving them simultaneously, we get:

y = 1.5 and x = 6

Therefore, the weight of the 2nd mixture required to make the third mixture is 6 litres.

Hence, the correct option is (c) 6 litres.