
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system

Find the maximum possible area of a section passing through the diagonal $XY$ of an arbitrary parallelepiped with edges $a \leq b \leq c$. The section is a parallelogram $Z X T Y$ with vertices lying on opposite edges of the parallelepiped. The area of the parallelogram equals the product of the length of the diagonal $XY$ and the distance from point $Z$ to $XY$. 

The largest area of a possible section, considering diagonal sections, is determined, with the given sides $a$, $b$, and $c$ in order of $a \le b \le c$. The areas of potential sections are:

$$
S_{1}=a \sqrt{b^{2}+c^{2}}, S_{2}=b \sqrt{a^{2}+c^{2}}, \text { and } S_{3}=c \sqrt{b^{2}+a^{2}} .
$$

Given the condition $a \leq b \leq c$, it is apparent that $S_{1} \leq S_{3}$ and $S_{2} \leq S_{3}$. Therefore, the maximum area section passes through the longest edge.

Thus, the maximum area section is $10 \sqrt{4^{2}+3^{2}}=50$.

Present the answer in LaTex format: \boxed{Your answer}
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