This is more of a λ-Calculus question (unrelated to combinators, really). But S takes 3, K takes 2, and I takes 1. This isn't strictly correct because S/K/I isn't a function, but rather a chain of lambdas with no free variables (and we get the number of "parameters the function takes" by counting the lambda symbols). See §1 in [1]. In your specific case,
S 11 12 = 12
K 11 12 13 = 11 13 (12 13)
I 11 = 11
Keep in mind that the literals "11," "12," and "13" in λ-Calculus aren't numbers.
> K takes 3 values, and returns the 1st, 3rd, and (2nd applied to 3rd)
Everything applies: so, e.g., when we see something like ABC, we actually omit the parentheses; if we include them, we have (((A)B)C). This can be confusing because applications associate to the left, but lambdas associate to the right[1].
So 11 13 (12 13) is actually (((11)13)((12)13)). In roundabout English, 11 applies to 13 which then applies to the result of 12 applies to 13.
S = lambda x: lambda y: lambda z: x(z)(y(z))
K = lambda x: lambda y: x
I = lambda x: x
That there is runnable code and with it you can implement numbers, math, logic, etc.
That's all the SKI combinators are: those functions. You apply them to themselves in various patterns and they model numbers and arithmetic and programming languages.
In [1]: S = lambda x: lambda y: lambda z: x(z)(y(z))
In [2]: K = lambda x: lambda y: x
In [3]: I = lambda x: x
In [4]: K(I)
Out[4]: <function __main__.<lambda>.<locals>.<lambda>(y)>
In [5]: _(S) is I
Out[5]: True
In [6]: K(I)(S) is I
Out[6]: True
I think you're definitely getting it, but you're also trying to see how these combinators are useful or at least usable. That's a bit of a tricky question because as @carapace mentions, you'd need to start combining them in interesting ways to get anything useful out of SKI.
[1] https://personal.utdallas.edu/~gupta/courses/apl/lambda.pdf