Determine the generator of an ideal of ring of integers

You seem to have touched upon several different ideas here.

Generators of the ideal $(3)$. Usually, when you talk about the generators of $(3)$, you mean its generators as an abelian group.

Define $\theta := \tfrac 1 2 (1 + \sqrt{-83})$. We know that the ring of integers $\mathbb Z[\theta]$ is generated by $\{1, \theta \}$ as an abelian group, so $(3)$ is generated by $\{3, 3\theta \}$ as an abelian group.

But having read the remainder of your question, it looks like this is not what you're after! What you're really interested in are the generators of the ideal class group for $\mathbb Z[\theta]$. (In fact, to prove that $\mathbb Z [\theta]$ is not a UFD, we don't even need to explicitly identify a set of generators for the ideal class group - it's enough to show that the ideal class group is non-trivial.)

Minkowski constant. The fact that $\frac 4 \pi \times \frac {2!} {2^2} \sqrt{83} \approx. 5.8$ implies that the ideal class group is generated by prime ideals that are factors of $(2)$, $(3)$ or $(5)$.

Dedekind's criterion. Dedekind's criterion is a way of factorising $(3)$ as a product of primes.

As you pointed out, we have the factorisation $$ x^2 - x + \frac {84}{2} \equiv x(x-1) \mod 3,$$

so Dedekind's criterion says that $$ (3) = (3, \theta)(3, \theta - 1)$$ is the prime factorisation of $(3)$ in $\mathbb Z[\theta]$.

Whether $\mathbb Z[\theta]$ is a UFD. This is equivalent to asking whether the ideal class group is trivial, i.e. whether all ideals are principal.

Why don't we check whether $(3, \theta)$ is principal? The answer must be "no". Note that $(3, \theta)$ has norm $3$. If it was principal, then there would exist $x, y \in \mathbb Z$ such that $(3, \theta) = (x + y\theta)$. But then $$ 3 = N(3, \theta) = N(x + y\theta) = (x + \tfrac 1 2 y)^2 + 83(\tfrac 1 2 y)^2,$$ which is impossible.

So $\mathbb Z[\theta]$ is not a principal ideal domain, and hence, is not a unique factorisation domain.

The minimal polynomial of $\alpha=\frac{1+\sqrt{-83}}{2}$ is $f = x^2 - x + 84/4 = x^2 - x + 21$. Modulo $3$ this factors as $$f \equiv x^2 - x = x(x-1) \pmod 3,$$ so by the Kummer-Dedekind theorem the ideal $(3)$ factors into primes in $\mathbb{Z}\bigg[\frac{1+\sqrt{-83}}{2}\bigg]$ as $(3) = (3, \alpha)(3, \alpha-1)$.

The ideal $(3)$ is principal, generated by $3$. You can show that the prime factors are not principal, e.g. using the field norm:

If $(3,\alpha) = (\beta)$ then $N(\beta)$ divides both $N(3) = 3^2$ and $N(\alpha) = f(0) = 21 = 3\cdot 7$, so $N(\beta) = 3$.

If we write $\beta = a+b\alpha$ then $$N(\beta)=(a+b\alpha)(a+b(1-\alpha)) = \ldots = a^2+ab + 21b^2.$$

The equation $a^2+ab + 21b^2 = 3$ is an ellipse in the $(a,b)$-plane without integral points, so there is no such $\beta$.