You can use inclusion-exclusion.
There are $3^{10}$ ways to pick 10 balls if the order does matter.
If at least one of the conditions fails, then:
- Not enough red balls; there are $2^{10}$ ways of picking with no red balls;
- Not enough green balls; $2^{10}$ ways with no green balls, $10\times 2^9$ with exactly one green ball;
- Not enough blue balls; $2^{10}$ with no blue balls, $10\times 2^9$ with exactly one, $\binom{10}{2}\times 2^8$ with exactly two.
If at least two of the conditions fail, then
- Not enough red and green balls: $1$ ways with no red or green balls at all; $10$ with no red and exactly one green ball.
- Not enough red and blue balls; $1$ way with no red or blue balls at all; $10$ with no red and exactly one blue; $\binom{10}{2}$ with no red and exactly two blue.
- Not enough green and blue; $1$ way with no green or blue; $10$ with no green and exactly one blue; $\binom{10}{2}$ with no green and exactly two blue; $10$ with exactly one green and no blue; $10\times 9$ ways with exactly one green and one blue; $10\times\binom{9}{2}$ with exactly one green and two blue.
You cannot have all three conditions fail.
So in summary, we have: $\begin{align*} 3^{10} &- (2^{10} + 2^{10}+10\times 2^9 + 2^{10}+10\times 2^9+45\times 2^{8}) \\ &+ (1 + 10 + 1 + 10 + 45 + 1 + 10 + 45 + 10 + 90 + 360)\\ &= 34800. \end{align*}$