Given a parallelepiped $A B C D A 1 B 1 C 1 D 1$. Points $M, L$, and $K$ are taken on the edges $A D, A 1 D 1$, and $B 1 C 1$ respectively, such that $B 1 K=\frac{1}{3} A 1 L, A M=\frac{1}{2} A 1 L$. It is known that $K L=2$. Find the length of the segment by which the plane $K L M$ intersects the parallelogram $A B C D$.
Let the secant plane intersect the line $B C$ at point $N$, and the line $A B$ at point $P$. By the theorem of the intersection of two parallel planes by a third, $M N \| K L$. Let $A 1 L=6 a$. Then $B 1 K=2 a, A M=3 a$. Draw lines through points $L$ and $K$ parallel to $A A 1$ until they intersect the lines $A D$ and $B C$ at points $L^{\prime}$ and $K^{\prime}$, respectively. (Points $L^{\prime}$ and $K^{\prime}$ are the parallel projections of points $L$ and $K$ onto the plane $A B C D$ with the projecting line $A A 1$). Then

$$
A L^{\prime}=A 1 L=6 a, B K^{\prime}=B 1 K=2 a, M L^{\prime}=A L^{\prime}-A M=6 a-3 a=3 a \text {, }
$$

$$
N K^{\prime}=M L^{\prime}=3 a, B N=N K^{\prime}-B K^{\prime}=3 a-2 a=a
$$

From the similarity of triangles $B P N$ and $A P M$, we find that

$$
\frac{M P}{P M}=\frac{B N}{A M}=\frac{a}{3 a}=\frac{1}{3}
$$

Therefore, $\frac{P M}{M N}=\frac{3}{1}$. Consequently,

$$
P M=\frac{3}{1} M N=\frac{3}{1} K^{\prime} L^{\prime}=\frac{3}{1} K L=\frac{3}{1} \cdot 2=\frac{3}{2} .
$$

## Answer

$\frac{3}{2}$.