
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

Let's conduct \( DL \perp AC \), \(LK \parallel CC_1\) (where \( K \in AC_1 \)), and \( PK \parallel DL \). By marking segment \( BQ = PD_1 \) on the lateral edge \( BB_1 \), we form a parallelogram \( PAQC_1 \), which will be the intersection of minimum area; here, \( AC_1 \) is the larger diagonal, and \( PQ \) is the smaller diagonal, both given in the problem statement.

Given:
\[ AC_1 = 3, \quad PQ = \sqrt{3}, \quad \varphi = 30^\circ \]
\[ \sin \varphi = \frac{1}{2}, \quad \cos \varphi = \frac{\sqrt{3}}{2} \]

Calculations:
\[ DL = PK = \frac{1}{2} \cdot PQ \cdot \sin \varphi = \frac{1}{2} \cdot \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{4} \]
\[ OK = \frac{1}{2} \cdot PQ \cdot \cos \varphi = \frac{1}{2} \cdot \sqrt{3} \cdot \frac{\sqrt{3}}{2} = \frac{3}{4} \]

\[ \frac{CL}{AL} = \frac{C_1 K}{AK} = \frac{C_1 O - OK}{C_1 O + OK} = \frac{3/2 - 3/4}{3/2 + 3/4} = \frac{1}{3} \]

Let \( CL = x \), then \( AL = 3x \), \( AC = 4x \).

\[ DL^2 = CL \cdot AL \] 
\[ \frac{3}{16} = 3x^2 \]
\[ x = \frac{1}{4}, \quad AC = 1 \]
\[ CC_1 = \sqrt{AC_1^2 - AC^2} = \sqrt{9 - 1} = 2\sqrt{2} \]

Volume \( V = AC \cdot DL \cdot CC_1 = 1 \cdot \frac{\sqrt{3}}{4} \cdot 2\sqrt{2} = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2} \).

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