The edge of a regular tetrahedron \(ABCD\) is \(a\). Points \(E\) and \(F\) are taken on the edges \(AB\) and \(CD\) respectively, such that the sphere circumscribed around the tetrahedron intersects the line passing through \(E\) and \(F\) at points \(M\) and \(N\). Find the length of the segment \(EF\), if \(ME: EF: FN = 3: 12: 4\).
Let $M E=3 t, E F=12 t, F N=4 t$. Let $P$ and $Q$ be the midpoints of edges $A B$ and $C D$ respectively. The segment connecting the midpoints of opposite edges of a regular tetrahedron is perpendicular to these edges, and its length is $\sqrt{2}$, where $a-$ is the length of the edge of the tetrahedron. Denote $P E=x, F Q=y$. By the theorem of the product of segments of intersecting chords of a sphere $A E \cdot E B=M E \cdot E N, C F \cdot F D=N F \cdot F M$, or

$$
\left(\frac{a}{\overline{2}-x)}\left(\frac{a}{2}+x\right)=3 t \cdot 16 t,\left(\frac{\pi}{2}-y\right)(\bar{a}+y)=4 t \cdot 15 t\right.
$$

$$
\frac{a^{2}}{4}-x 2=48 t 2, \frac{a^{2}}{4}-y 2=60 t 2
$$

from which $x 2+y 2=\frac{n^{2}}{2}-108 t 2$. From the right triangles $E P Q$ and $F Q P$ we find that

$$
144 t 2=E F 2=E P 2+P F 2=E P 2+(P Q 2+Q F 2)=x 2+\frac{\frac{0}{2}^{2}}{2}+y 2
$$

Then $144 t 2=\frac{a^{2}}{2}-108 t 2+\frac{a^{2}}{2}$, from which $t 2=\frac{a^{2}}{252}, t=\frac{a}{6 \sqrt{7}}$. Therefore, $E F=12 t=\frac{\frac{2 a}{\sqrt{7}}}{\sqrt{2}}$.

## Answer

$\frac{2 a}{\sqrt{7}}$