Let the position vectors of the vertices of tetrahedron \(O A B C\) are \(\mathbf{O A}=\mathbf{a}, \mathbf{O B}=\mathbf{b}\) and \(\mathbf{O C}=\mathbf{c}\)
So, volume of tetrahedron \(O A B C=\frac{1}{6}[\mathbf{a} \mathbf{b ~ c}]\)
Now, position vectors of vertices of new tetrahedron are \(\frac{\mathbf{a}+\mathbf{b}}{3}, \frac{\mathbf{b}+\mathbf{c}}{3}, \frac{\mathbf{c}+\mathbf{a}}{3}\) and \(\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}\),
so, coterminous edge vectors of the new tetrahedron are \(\frac{\mathbf{a}}{3}, \frac{\mathbf{b}}{3}\) and \(\frac{\mathbf{c}}{3}\).
\(\therefore\) Volume of new tetrahedron is \(\frac{1}{6}\left[\frac{\mathbf{a}}{3} \frac{\mathbf{b}}{3} \frac{\mathbf{c}}{3}\right]\)
\(=\frac{1}{6 \times 27}[\mathbf{a ~ b c}]\)
So, the required ratio
\(=\frac{1}{27}\)
Hence, option (2) is correct.