A group with at least $6k$ elements with the property that among any $3k$ of its elements $k$ of them are in $Z(G)$

Question

$\DeclareMathOperator{\ord}{ord}$ Let $(G,\cdot)$ be a finite group with $\ord G \ge 6k$ ($k\in \mathbb{N}$) such that among any $3k$ of its elements there are $k$ elements from $Z(G)$. Prove that $G$ is abelian.
I will present my reasoning for $k\ge 2$.
$\textbf{Lemma:}$ If $(G,\cdot)$ is a non-abelian group with $n$ elements, then $|Z(G)|\le \frac{n}{4}$.
$\textbf{Proof:}$ Let $a\in G\setminus Z(G)$. We have that $C(a)\neq G$. Hence, $C(a)$ is a proper subgroup of $G$.
From Lagrange's theorem it follows that $|C(a)| \le \frac{|G|}{2}$. Since $a\notin Z(G)$ and $a\in C(a)$ we have that $Z(G)\neq C(a)$, so $Z(G)$ is a proper subgroup of $C(a)$.
From Lagrange's theorem we deduce that $|Z(G)|\le \frac{|C(a)|}{2}\le \frac{|G|}{4}$.
Now let's get back to the problem. We will assume that $G$ is not abelian.
Obviously, $\exists t\in \mathbb{N}, t\ge 2$ and $r\in \{0,1,...3k-1\}$ such that $\ord G=3kt+r$.
Suppose that $|G\setminus Z(G)|\ge 3k-1$. Then, $|(G\setminus Z(G)\cup\{e\}|\ge 3k$, so in this set there are $k$ elements from $Z(G)$, which means that there is at least an element from $Z(G)$ in $G \setminus Z(G)$. We have reached a contradiction.
So, we have that $|G \setminus Z(G)|\le 3k-2$.
It now follows that $|Z(G)|=|G|-|G\setminus Z(G)|\ge 3kt+r-3k+2$.
We will prove that $3kt+r-3k+2> \frac{3kt+r}{4}\iff 9kt-12k+3r+8 >0$, which is obviously true and we have reached a contradiction to our Lemma.
Hence, our assumption was wrong and $G$ is abelian.
Now, for the case $k=1$ we only need to study what happens when $\ord(G) \in \{6, 7, 8, 9, 10, 11 \}$. This is pretty easy and I will not include it here since it just involves some computations.
I would like you to tell me if my solution is correct and if you have any better approach to this.

Answer 1

Hint for a quick proof:

Take $\;3k\;$ elements, thus $\;k\;$ elements (at least) of these are in $\;Z(G)\;$. Now take the remaining $\;3k\;$ elements of the group, and again we have another, different, at least $\;k\;$ elements in $\;Z(G)\;$. Thus, we already got that $\;|Z(G)|\ge2k\;$, and thus

$$\left|G/Z(G)\right|\le\frac{6k}{2k}=3\implies G/Z(G)\;\;\text{ is cyclic}\ldots$$

Now finish the proof.