Distributing 7 red balls and 2 blue balls into 3 containers such that in each container, there are 3 balls and at least 2 of them are red balls. How many ways are there to distribute 7 distinguishable red balls and 2 distinguishable blue balls into 3 indistinguishable containers such that in each container, there are 3 balls and at least 2 of them are red balls?

I got $\dfrac{\binom{7}{2}\cdot\binom{5}{2}\cdot\binom{3}{2}\cdot 3\cdot 2}{3!}=630$, because there are $\binom{7}{2}\cdot\binom{5}{2}\cdot\binom{3}{2}$ to distribute 6 red balls so each container has 2 balls. Next, there are $3$ containers for the last 7 ball and 2 ways to distribute the 2 blue balls. Lastly, divide by $3!$ because the containers are indistinguishable. However, this is the wrong answer. Can someone tell me what's wrong with my solution?
Thanks!

For another view point :
As @trueblueanil said , the only possible distribution can be done when one boxes has $3$ red balls and the other two take $2$ red balls. The blue balls will go directly to complement them to $3$ balls
To make our calculation easier , think that the boxes are distinct in the beginning.
Step 1-) Determine which box will have $3$ red balls by $C(3,1)$.
Step 2-) Distribute red balls to $3$  distinct boxes such that one box has three ball , the others have two balls by $C(7,3)\times C(4,2) \times C(2,2)$
Step 3-) Distribute $2$ blue balls to those boxes which have $2$ red balls such that each box will take $1$ blue balls ,so we can do it by $C(2,1) \times C(1,1)$.
Now we found the number of distributing $7$ distinct red balls and $2$ distinct blue balls to $3$ distinct boxes such that one box has $3$ red balls and the other two boxes have $2$ red balls and $1$ blue ball. It is equal to $$C(3,1)\times C(7,3)\times C(4,2) \times C(2,2) \times C(2,1) \times C(1,1)=1260$$
However, we know that those three boxes were identical , so divide the result by $3!$ to convert distinct boxes into identical.
$$\frac{C(3,1)\times C(7,3)\times C(4,2) \times C(2,2) \times C(2,1) \times C(1,1)}{3!}=210$$
