Worst case, median case, and best case complexity have nothing to do with O/Θ/Ω.

When analyzing an algorithm, you want to compute the number of steps it has to execute in certain cases as a function of its input size(s). That function will be something like 2n^2 + 67n + 189 steps. Since this is too much information, you can then express it instead as O(n^2) (or O(n^100)) or Ω(n^2) (or Ω(25n) ) or Θ(n^2).

You can then also compute the best case running time of the algorithm as a function of the size of the input, and get 2n^2 + 67n. Then you can say that the best case running time is also O(n^2).

Finally, you can (sometimes) define an "average" running time.

Now, what do we mean by "shortest running time"? Is that not always for n=0 or something? No - the best case running time has to do with the running time in a case where the input has a certain shape. For some sorting algorithms for example, if the input is already sorted, they will only go through the list once and do nothing, so they have a best case running time that is O(n). Other sorting algorithms will still perform steps even if the input is already sorted, so even their best case running time may be O(n log(n)).

Now, you can say that, given any possible input, the running time of the algorithm is O(worse case running time) and Ω(best case running time). You definitely CAN'T say in general that it is Θ(average case running time) though, as in will sometimes run in time that is shorter than "average case running time" (for best case scenarios); or sometimes it will run in time that is longer than "average case running time" (for worse case scenarios).

Basically, the worse/best/average case running times are an analysis of how properties other than the size of the input affect its running time.

Big-O notation and its friends have nothing to do with complexity analysis. They are a general notation for the asymptotic behavior of mathematical functions. Big-O is usually understood as an upper bound on a function, big-Ω is a lower bound, and Θ is a tight bound (the function f is by definition always within a factor of k above or below Θ(f)).

Edit: let's take an example. Say we have the following algorithm:

isAllOdd(array x):
if x is empty:
return true
if x[0] is even:
return false
return isAllOdd(x[1:])

In the best case, for any size array with the first element being even, this function finishes after executing 4 instructions: 2 comparisons, an array access and then a return. So it's best case complexity is best(n) = 4, which is O(1), Θ(1), Ω(1).

In the worse case, all elements of the array are odd, so we have to go through all n elements to find that out. It will have to execute 6 instructions for each element in the array (2 comparisons, 2 array accesses, a call and a return), except for the last element where it will execute only 2 instructions (check size then return). So its worst case complexity function is worst(n) = 6(n-1) + 2, which is O(n), Θ(n), Ω(n).

In the average case, for every call we are flipping a coin, and we have n total flips. Since we can assume the coin is fair, we will expect to see an even number after log(n/2) calls of isAllOdd. So, our average case complexity is average(n) = 6*log((n-1)/2) + 2 steps, which is O(log n), Θ(log n), Ω(log n). (I am ignoring cases where n < 2, which is anyway irrelevant for asymptotic behavior).

Now, what is the overall complexity of isAllOdd? We can generalize the average case calculation if we want to get an exact value; but whatever that function, overall(n), is, we know for sure best(n) < overall(n) < worst(n); so overall(n) is O(n) and Ω(1). However, we can't define any Θ for it - there is no function which it will differ from by at most a constant factor for any n.

Why do the detailed analysis instead of saying this is O(n) and being done with it? Well, perhaps we know that we are in fact calling this function with a large number of even numbers, so we would actually over-estimate its complexity. If we instead know we are calling it with a large number of odd numbers, we would also find out from the average case analysis that it would be better to define a reverse funtion, isAllEven, and compute isAllOdd = !isAllEven.