$[\underline{\text { Properties of Sections }}]$

The edge of the cube $A B C D A 1 B 1 C 1 D 1$ is 12. Point $K$ lies on the extension of edge $B C$ at a distance of 9 from vertex $C$. Point $L$ on edge $A B$ is 5 units away from $A$. Point $M$ divides the segment $A 1 C 1$ in the ratio $1: 3$, counting from $A 1$. Find the area of the section of the cube by the plane passing through points $K, L, M$.
Let the line $L K$ intersect the edge $C D$ at point $F$ (Fig.1). According to the theorem on the intersection of two parallel planes by a third intersecting plane, the plane intersects the base $A 1 B 1 C 1 D 1$ along a line $a$, passing through point $M$ parallel to $L F$. Let the line $a$ intersect the lines $A 1 B 1$ and $C 1 D 1$ at points $N$ and $E$ respectively, and let points $L 1$, $F 1$, and $K 1$ be the orthogonal projections of points $L$, $F$, and $K$ onto the plane $A 1 B 1 C 1 D 1$. Denote $\angle B K L = \angle B 1 K 1 L 1 = \alpha$. Then

$$
\operatorname{tg} \alpha = \frac{B L}{B K} = \frac{7}{21} = \frac{1}{3}, \quad C 1 F 1 = C F = C K \operatorname{tg} \alpha = 9 \cdot \frac{1}{3} = 3, \quad \cos \alpha = \frac{3}{\sqrt{10}}
$$

Let $F 1 E = L 1 N = x$ (Fig.2). Then

$$
A 1 N = A 1 L 1 - L 1 N = A L - L 1 N = 5 - x, \quad C 1 E = C 1 F + F 1 E = 3 + x,
$$

$$
\frac{A_{1} N}{C_{1} E} = \frac{A_{1} M}{M C_{1}} = \frac{1}{3}, \quad \frac{5 - x}{3 + x} = \frac{1}{3}
$$

From this, we find that $L 1 N = x = 3$. Then $C 1 E = 6$, $A 1 N = 2$. Points $E$ and $N$ lie on the sides $C 1 D 1$ and $A 1 B 1$. Therefore, the section of the given cube by the plane passing through points $K, L$, and $M$ is a parallelogram $L F E N$. Drop a perpendicular $L H$ from point $L$ to the line $NE$ (Fig.1). Then $L H$ is the height of the parallelogram $L F E N$. By the theorem of three perpendiculars, $L 1 H \perp N E$. From the right triangle $L 1 H N$ we find (Fig.2) that

$$
L 1 H = L 1 N \cos \angle N L 1 H = L 1 N \cos \alpha = 3 \cdot \frac{3}{\sqrt{10}} = \frac{9}{\sqrt{10}}
$$

Thus,

$$
L H = \sqrt{L_{1} L^{2} + L_{1} H^{2}} = \sqrt{144 + \frac{81}{10}} = \frac{39}{\sqrt{10}}
$$

Drop a perpendicular $F P$ to $A B$. Then

$$
\begin{gathered}
L P = B L - B P = 7 - 3 = 4 \\
N E = L F = \sqrt{F P^{2} + L P^{2}} = \sqrt{144 + 16} = 4 \sqrt{10}
\end{gathered}
$$

Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_b09eb9152e3973547418g-14.jpg?height=120&width=602&top_left_y=2101&top_left_x=728)

## Answer

156.00