Let $S$ be a set of 2001 distinct positive integers from the set $\{1, 2, 3, \dots, 3000\}$. Let us look at 500 sets
$$K_j = \{6j + i \mid i = 1, 2, 3, 4, 5, 6\}, \quad j = 0, 1, 2, \dots, 499.$$
There is a set with at least five elements of $S$ among them (since $4 \cdot 500 < 2001$).
If three of these five elements are odd numbers, we found the required triple.
Otherwise, the set $K_j$ contains three even and two odd numbers from $S$.
These two odd numbers are relatively prime since their difference is 2 or 4.
Three even numbers from the same $K_j$ are three consecutive even numbers. Only one of them is divisible by 3 and at most one of them can be divisible by 5. Therefore we can choose an even number which is not divisible by 3 nor by 5.
That even number together with two odd numbers gives the required triple. Since the difference of the numbers from the set $K_j$ is less than 6, only one of them can be divisible by prime $p$, $p \ge 7$.