3. In a rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$, the sides $A B=2, A C=$ 3, $A A_{1}=$ 4. Find the area of the section $A M K$, where $M$ is the midpoint of $B B_{1}$ and $K$ is the midpoint of $D D_{1}$.
Answer: $2 \sqrt{22}$

Solution. Let's construct the desired section (see figure). Point $P$ is the intersection of $A M$ and $A_{1} B_{1}$. Due to the position of point $M$, $A_{1} P$ is twice as large as $A_{1} B_{1}$. Similarly, we construct point $Q$, where $A_{1} Q = 2 A_{1} D_{1}$. Therefore, $P Q$ is parallel to $B_{1} D_{1}$ and passes through point $C_{1}$. Thus, the section is a parallelogram $A K C_{1} M$.

The area can be calculated in different ways.

1st method. Notice that $K M = \sqrt{4+9} = \sqrt{13}$, $A M = \sqrt{4+4} = 2 \sqrt{2}$, and $A K = \sqrt{9+4} = \sqrt{13}$. Therefore, triangle $AKM$ is isosceles, and its height is $\sqrt{13-2} = \sqrt{11}$. Thus, the area of the section is $2 \sqrt{2} \cdot \sqrt{11} = 2 \sqrt{22}$.

2nd method. Use vectors. It is known that $S = |\overrightarrow{A K}| \cdot |\overrightarrow{A M}| \cdot \sin \varphi$, while $(\overrightarrow{A K} \cdot \overrightarrow{A M}) = |\overrightarrow{A K}| \cdot |\overrightarrow{A M}| \cdot \cos \varphi$, so $S^{2} = \overrightarrow{A K}^{2} \cdot \overrightarrow{A M}^{2} - (\overrightarrow{A K} \cdot \overrightarrow{A M})^{2}$. Notice that a coordinate system can be associated with the parallelepiped, with the origin at point $A$. We have $\overrightarrow{A K} = (0, 3, 2)$ and $\overrightarrow{A M} = (2, 0, 2)$. Then $\overrightarrow{A K}^{2} = 3^{2} + 2^{2} = 13$, $\overrightarrow{A M}^{2} = 2^{2} + 2^{2} = 8$, and $(\overrightarrow{A K} \cdot \overrightarrow{A M}) = 0 + 0 + 4 = 4$, from which the answer follows.

Remark. Other methods of calculating the area are possible. For example, using the cross product of vectors $\overrightarrow{A K}$ and $\overrightarrow{A M}$.

Criteria. In the case of full justification - a score of 7 points.