If, according to 2's Complement any binary string can be converted to it's negative counterpart by flipping each digit to it's opposite number (eg. $1001_2 \to 0110_2$) and then add 1, then how would you implement a "3's Complement"? So to speak.
Like, lets say I have $29_{10}$, and want to write it as $-29_{10}$ in binary.
First I convert it into binary: $11101_2$. And now to make it negative, we use 2's Complement. We change the binary number to $00010_2$ and add $1 \to 00011_2$.
Now we have $-29_1$$_0$ in binary form.
But now I want to turn $29_1$$_0$ into negative ternary number.
So, again we convert to it's ternary form: $1002_3$
Therefore you can probably guess that I now need to convert it to negative; Is there any way that I can apply both 10's Complement and 2's complement to create a sort of "3's Complement?
You can do this is any base. For example, in base 10, you can have "$9'$s complement." The old rotary mechanical adding machines worked this way. "Two's complement" is a not really a good name for the system used on computers, since flipping the bits means subtracting from one, but "one's complement" was already used for an older system.
In any case, for ternary, subtract each digit from $2$ and add $1$. So assuming you're using 4-digit integers, $$1002_3\to1220_3\to\boxed{1221_3}.$$ The idea is very simple. If you add $1$ to $2222_3$ you get $0000_3$ assuming the carry out of the last place is lost, so you just need to subtract from $2222_3$ and add $1$ to the result.