A hypergraph $H$ consists of a finite vertex set $V(H)$ of elements, called vertices, together with a finite multiset $E(H)$ of subsets of $V(H)$, called hyperedges. The order of $H$ is $n(H)=|V(H)|$ and the size of $H$ is $m(H) = |E(H)|$. The hypergraph $H=(V,E)$ is called a $k$-uniform hypergraph if all hyperedges in $H$ are of the same arity $k$. The degree of a vertex $v$ in $H$, denoted by $d_H(v)$, is the number of hyperedges of $H$ that contain $v$. The hypergraph $H$ is called $r$-regular if $d_H(v) = r$ for every vertex $v$ in $H$. A subset of the vertices of a hypergraph $H$ is a transversal of $H$ if it intersects every hyperedge of $H$. The transversal number of a hypergraph $H$ is the minimum cardinality of a transversal, denoted by $\tau(H)$. In other words, the transversal number of $H$ is the minimum number of vertices needed to meet every hyperedge. \par In 2002, Z. Tuza and P.~D. Vestergaard [Discuss. Math. Graph Theory {\bf 22} (2002), no.~1, 199--210; MR1936237] posed the following conjecture: \par {\it Tuza-Vestergaard Conjecture}. If $H$ is a 3-regular 6-uniform hypergraph of order $n$, then $\tau(H) \leq \frac{1}{4n}$. \par After 22 years, the authors prove the correctness of the Tuza-Vestergaard conjecture in the paper under review. In order to prove that the Tuza-Vestergaard conjecture holds, they prove the following key result. \par Theorem 1. Let $\Cal{H}_{3,6,n,m}$ be the class of all 6-uniform hypergraphs of order $n$ and size~$m$ with maximum degree at most 3. If $H\in \Cal{H}_{3,6,n,m}$ then $\tau (H)\leq \frac{1}{6} (n +m)$. \par Now suppose $H$ is a 3-regular 6-uniform hypergraph of order $n$ and size $m$; then ${H\in \Cal{H}_{3,6,n,m}}$ and $2m=n$. So, by Theorem 1, it follows that $\tau(H) \leq \frac{1}{4n}$. \par Theorem 2 (Tuza-Vestergaard Theorem). If $H$ is a 3-regular 6-uniform hypergraph of order $n$, then $\tau(H) \leq \frac{1}{4n}$.