In a group of 3 girls and 4 boys, there are two boys $$B_{1}\text{ and }B_{2}$$. The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but $$B_{1}\text{ and }B_{2}$$ are not adjacent to each other, is :

Let's assume all girls together as 1 block (G) and all boys together as another block (B).

So we have 2 blocks: G and B. They can be arranged in:  2! = 2 ways

Arrange the girls inside their block = 3! = 6 ways

Arrange boys inside their block = 4! = 24 ways

Now, we can subtract the cases where B1 and B2 are adjacent. 

If we assume B1 and B2 as one unit, then we have 3 units of boys. They can be arranged in 3! ways, and B1 and B2 can also arrange themselves in two ways. 

So, adjacent cases = 3! × 2 = 6 × 2 = 12

Valid arrangements: 24 - 12 = 12

Hence, the total number of ways: 2 × 6 × 12 = 144