Reducing 3-coloring problem to trio representatives
Let $G = (V,E)$ be an instance of $3$ coloring. Construct an instance $\phi$ of 3-SAT as follows.


For each vertex $v \in V$ create $3$ variables $v_a, v_b, v_c$ representing the $3$ possible ways to color of $v$.
For each vertex $v \in V$ create the clauses $(v_a \vee v_b \vee v_c) \wedge (\overline{v_a} \vee \overline {v_b}) \wedge (\overline{v_b} \vee \overline {v_c}) \wedge (\overline{v_c} \vee \overline {v_a})$. This encodes the fact that $v$ must be colored by exactly one color.
For each edge $(u,v) \in E$ create the clauses $( \overline{v_a} \vee \overline{u_a} ) \wedge ( \overline{v_b} \vee \overline{u_b} ) \wedge ( \overline{v_c} \vee \overline{u_c} )$. This ensures that $u$ and $v$ cannot be given the same color.


Clearly the above reduction can be performed in poylnomial-time and ensures that there is a satisfying assignment to $\phi$ iff there is a 3-Coloring for $G$.

Now transform the instance $\phi$ of 3-SAT into an equivalent of instance $\phi'$ of 1-in-3 SAT  (see here for the definition of 1-in-3 sat and the reduction from 3-SAT).

We have hence shown that 3-Coloring is polynomial-time reducible to 1-in-3 SAT.
It turns out that 1-in-3 SAT is exactly your problem: the set of students is the set of variables and, for each clause $\{x_i, x_j, x_k\}$ you create a group containing students $x_i$, $x_j$, and $x_k$. Setting a variable to true corresponds to selecting the corresponding student as a leader.