[ Coordinate method in space ] [

The edge of the cube $E F G H E 1 F 1 G 1 H 1$ is 2. Points $A$ and $B$ are taken on the edges $E H$ and $H H 1$, respectively, such that $\frac{E A}{A H}=2, \frac{B H}{B H 1}=\frac{1}{2}$. A plane is drawn through points $A, B$, and $G 1$. Find the distance from point $E$ to this plane.

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## First II method.

Since $\frac{H A}{H E}=\frac{H B}{H H_{1}}$, the line $A B$ is parallel to the diagonal $H 1 E$ of the square $E H H 1 E 1$, and since $H 1 E \| G 1 F$, then $A B \| G 1 F$. Therefore, the lines $A B$ and $G 1 F$ lie in the same plane - the cutting plane passing through points $A, B$ and $G 1$. Thus, the trapezoid $A B G 1 F$ is the section mentioned in the problem. Let $K$ be the intersection point of $A F$ and $E G$. Then the cutting plane and the plane of the diagonal section $E G G 1 E 1$ intersect along the line $G 1 K$. The line $G 1 K$ intersects the line $E E 1$ lying in the same plane $E G G 1 E 1$ at some point $D$. Then the required distance from point $E$ to the plane $A B G 1 F$ is the height of the tetrahedron $D A F E$, drawn from vertex $E$. From the similarity of triangles $A K E$ and $F K G$, it follows that

$$
\frac{K E^{\prime}}{K G}=\frac{A E}{F G}=\frac{2 H E}{F G}=\frac{2}{3}
$$

From the similarity of triangles $D K E$ and $G 1 K G$, we find that

$$
D E=G G 1 \cdot \frac{D K}{K G_{1}}=G G 1 \cdot \frac{K E}{K G}=\frac{2}{3} \cdot 2=\frac{4}{3}
$$

Then

$$
\begin{gathered}
D F=\sqrt{D E^{2}+E F^{2}}=\sqrt{\left(\frac{4}{3}\right)^{2}+2^{2}}=\frac{2 \sqrt{13}}{3}, \\
A F=\sqrt{A E^{2}+E F^{2}}=\sqrt{\left(\frac{4}{3}\right)^{2}+2^{2}}=\frac{2 \sqrt{13}}{3} \\
A D=A E \sqrt{2}=\frac{4 \sqrt{2}}{3}
\end{gathered}
$$

Let FP be the height of the isosceles triangle $A D F$. Then

$$
F P=\sqrt{A F^{2}-\frac{1}{4} A D^{2}}=\sqrt{\frac{52}{9}-\frac{8}{9}}=\sqrt{\frac{44}{9}}=\frac{2 \sqrt{11}}{3}
$$

Let $V$ be the volume of the tetrahedron $D A F E$. Since $A E$ is the height of the tetrahedron, then

$$
V=\frac{1}{3} S_{\triangle D E F} \cdot A E=\frac{1}{3} \cdot \frac{1}{2} D E \cdot E F \cdot A E=\frac{1}{6} \cdot \frac{4}{3} \cdot 2 \cdot \frac{4}{3}=\frac{16}{27} .
$$

On the other hand, $V=\frac{1}{\overline{3}} S_{\Delta A D F} \cdot h$, where $h-$ is the required distance. Therefore,

$$
h=\frac{3 V}{S_{\triangle A D F}}=\frac{\frac{16}{2}}{\frac{1}{2} A D \cdot F^{\prime} P}=\frac{\frac{16}{3}}{\frac{1}{2} \cdot \frac{1 \sqrt{3}}{3} \cdot \frac{2 \sqrt{11}}{3}}=2 \sqrt{\frac{2}{11}}
$$

## Second method.

Introduce a coordinate system $O x y z$, directing the axis $O x$ along the ray $H G$, the axis $O y$ - along the ray $H E$, the axis $O z$ - along the ray $H H 1$.

Find the coordinates of the required points: $G 1(2 ; 0 ; 2), E(0 ; 2 ; 0)$. Write the equation of the plane $A B G 1$ in segments:

$$
\frac{x}{a}+\frac{\frac{y}{2}}{3}+\frac{z}{2}+\frac{2}{3}=1
$$

Substituting the coordinates of point $G 1$ into this equation, we find that $a=-1$. Then the equation of the cutting plane is

$$
-x+\frac{3}{2} y+\frac{3}{2} z=1 \text {, or } 2 x-3 y-3 z+2=0 \text {. }
$$

Let $h$ be the required distance. By the formula for the distance from a point to a plane, we find that

$$
h=\frac{|2 \cdot-3 \cdot-2-3 \cdot-3+2|}{\sqrt{2^{2}+3^{2}+3^{2}}}=\frac{4}{\sqrt{22}}=2 \sqrt{\frac{2}{11}}
$$

## Answer

$2 \sqrt{\frac{2}{11}}$