[ $\quad$ Construction of sections $\quad]$

Each edge of the tetrahedron $A B C D$ is equal to 1. Point $E$ is on edge $A B$, point $F$ is on edge $B C$, and point $G$ is on edge $C D$ such that $A E=\frac{2}{3}, B F=\frac{3}{4}$, and $C G=\frac{1}{2}$. The plane $E F G$ intersects the line $A D$ at point $H$. Find the perimeter of triangle $H E G$.
Note that $A B C D$ is a regular tetrahedron. Consider the plane of triangle $A B C$. Let lines $E F$ and $A C$ intersect at point $T$. Draw a line through vertex $B$ parallel to $A C$. Let this line intersect line $E F$ at point $M$. Denote $C T=x$. From the similarity of triangles $B M F$ and $C F T$, we find that $B M = C T \cdot \frac{B F^{\prime}}{F C} = 3 x$. From the similarity of triangles $B E M$ and $A E T$, we find that

$$
1 + x = A T = B M \cdot \frac{A E}{E B} = 3 x \cdot 2 = 6 x
$$

from which $C T = x = \frac{1}{5}$. Consider the plane of triangle $A D C$. Draw a line through vertex $D$ parallel to $A C$. Let this line intersect line $T G$ at point $N$. Since line $T G$ lies in the cutting plane, the point of its intersection with line $A D$ is point $H$. From the equality of triangles $D G N$ and $C G T$, we find that $N D = C T = \frac{1}{5}$. From the similarity of triangles $D H N$ and $A H T$, it follows that

$$
\frac{D H}{A H} = \frac{D N}{A I^{\prime}} = \frac{1}{V} = \frac{1}{6}
$$

Therefore, $D H = \frac{1}{7}, A H = \frac{6}{7}$. Using the cosine theorem in triangles $D G H$ and $A E H$, we find that

$$
\begin{aligned}
& G H = \sqrt{D G^{2} + D H^{2} - 2 D G \cdot D H \cos 60^{\circ}} = \sqrt{\frac{1}{4} + \frac{1}{49} - \frac{1}{14}} = \frac{\sqrt{39}}{14} \\
& E H = \sqrt{A E^{2} + A H^{2} - 2 A E \cdot A H \cos 60^{\circ}} = \sqrt{\frac{4}{9} + \frac{36}{49} - \frac{4}{7}} = \frac{2 \sqrt{67}}{21}
\end{aligned}
$$

Let $O$ be the center of the base $A B C$ of the regular tetrahedron $A B C D$, and $K$ be the midpoint of side $A B$. Then $D O$ is the height of the tetrahedron, orthogonal projection of $G 1$ of point $G$ on the base plane lies on line $O C$, and $G 1$ is the midpoint of $O C$. Further, we have:

$$
\begin{gathered}
D O = \sqrt{\frac{2}{3}}, G G 1 = \frac{1}{2} D O = \frac{1}{2} \sqrt{\frac{2}{3}}, C K = \frac{\sqrt{3}}{2}, O C = \frac{\sqrt{3}}{3}, C G 1 = \frac{\sqrt{3}}{6} \\
G 1 K = C K - C G 1 = \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{3}, K E = A E - A K = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}
\end{gathered}
$$

$$
\begin{gathered}
G 1 E = \sqrt{K E^{2} + G_{1} K^{2}} = \sqrt{\frac{1}{36} + \frac{1}{3}} = \frac{\sqrt{13}}{6} \\
E G = \sqrt{G_{1} E^{2} + G_{1} G^{2}} = \sqrt{\frac{13}{36} + \frac{1}{6}} = \sqrt{\frac{19}{36}} = \frac{\sqrt{19}}{6}
\end{gathered}
$$

Therefore,

$$
G H + E H + E G = \frac{\sqrt{39}}{14} + \frac{2 \sqrt{67}}{21} + \frac{\sqrt{19}}{6}
$$

## Answer

$\frac{\sqrt{39}}{14} + \frac{2 \sqrt{67}}{21} + \frac{\sqrt{19}}{6}$