We can choose $c=3$ and $n_0=1$ for the definition of Big-Oh to hold.

$$3n^2 - 100n + 6 \le 3n^2$$

Recall the choice of $n_0$ and $c$ are not unique.

There can be many (actually, infinitely many) different combinations of $n_0$ and $c$ that would make the Big-Oh definition to work. It depends on what inequalities you use while doing the upper-bounding.

We can choose $c=1$ and $n_0=1$ for the definition of Big-Oh to hold.

$$3n^2 - 100n + 6 \le n^3$$

Big-Oh expresses an upper bound but does not necessarily provide a tight upper bound.

For example, you can easily show $T(n) \in \Omicron(n^p)$ for any $p \ge 2$.

There simply is no $c$ and $n_0$ where the Big-Oh definition would hold for $T(n) \in \Omicron(n)$.