I don't know if my answer is correct.
10P3 = 720 for the one person,
7P4 = 840 for the two person,
3P1 = 3 for the three person
720+840+3 = 1563 ways
Assume all 6 people are distinct as are all 10 prizes and we don't care what order they received the prizes, only which ones they get.
Your calculation is not correct in various respects ...
Assuming the winners are distinct:
Pick $3$ prizes for $1$ winner: You can pick $6$ different winners, and you can pick $3$ gifts in ${10 \choose 3} =120$ ways. You need to divide your $720$ by $3$ since, for example: picking gifts 1,4, and 7, is the same as picking gifts 4,7,and 1.
Now we need to pick two winners of $2$ prizes. We can pick those winner in ${5 \choose 2} = 10$ ways. Once we have done that, we can pick $2$ gifts from the remaining $7$ for the 'older' of the two in ${7 \choose 2} = 21$ ways, and then we can pick $2$ gifts from the remaining $5$ gifts for the 'younger' of the two in ${5 \choose 2} = 10$ ways.
Finally, the last $3$ prizes can only be divided in $3!=6$ ways between the last $3$ winners, rather than in just $3$ ways, which is what you said.
And, to get the grand total of possibilities, your big mistake is that you need to multiply those possibilities (since for each of the ways to pick the first group of $3$, there are the many ways to pick the next group of $2$, etc.). So, rather than add them, you should get:
$$6 \cdot {10 \choose 3} \cdot 10 \cdot {7 \choose 2} \cdot {5 \choose 2} \cdot 3! = 9072000$$