There are three vessels of equal capacity .Vessel A contains Milk and

Here is my approach, very similar to fluke's, just a different thought process of arriving at the same calculation.

The three vessels are of equal capacity (which we assume to be full since it is a PS question. I would prefer clearer direction though). The milk:water ratio is given for each as 4:3, 2:1 and 3:2. The reason we cannot just add the milk parts (4+2+3) and water parts (3+1+2) together is that they represent different fractions of the whole volume i.e. in the first vessel, milk is 4 parts out of a total of 7 parts while in the second vessel, milk is 2 parts out of a total of 3 parts etc. If we make the total number of parts in each ratio equal, then we can just add the 'milk parts' together and all the 'water parts' together.
Ratios:
4:3 (Total 7 parts)
2:1 (Total 3 parts)
3:2 (Total 5 parts)
How can we make the total number of parts equal in the 3 cases? By taking the LCM.
4*15 : 3*15 = 60:45 (Total parts 105)
2*35 : 1*35 = 70:35 (Total parts 105)
3*21 : 2*21 = 63:42 (Total parts 105)
When we mix the three solutions, the total number of parts of milk is 60+70+63 = 193
the total number of parts of water = 45+35+42 = 122
The required ratio = 193:122