Cross Products of Side Vectors of Regular Hexagon

To determine the validity of the statements, let us analyze the geometry of the regular hexagon $PQRSTU$.

Analysis of Statement-2:
In a regular hexagon $PQRSTU$, the opposite sides are parallel:

• Side $PQ$ is opposite and parallel to side $TS$. Therefore, the vectors along these sides are related by:
  $\vec{ST} = -\vec{PQ}$
  Since $\vec{PQ}$ and $\vec{ST}$ are collinear (anti-parallel), their cross product is:
  $\vec{PQ} \times \vec{ST} = \vec{0}$

• Side $RS$ is not parallel to side $PQ$; the angle between the vectors $\vec{PQ}$ and $\vec{RS}$ is $120^\circ$. Thus:
  $\vec{PQ} \times \vec{RS} \neq \vec{0}$

Statement-2 asserts that $\vec{PQ} \times \vec{RS} = \vec{0}$ and $\vec{PQ} \times \vec{ST} \neq \vec{0}$, which is directly contrary to the geometric properties established above. Therefore, Statement-2 is False.

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Analysis of Statement-1:
Using vector addition, we have:
$\vec{RS} + \vec{ST} = \vec{RT}$
In $\triangle RST$, the side lengths are equal ($RS = ST$) and the interior angle is $\angle RST = 120^\circ$.
Therefore, $\triangle RST$ is an isosceles triangle with:
$\angle STR = \angle SRT = \frac{180^\circ - 120^\circ}{2} = 30^\circ$

Since the line containing segment $ST$ is parallel to the line containing segment $PQ$, the vector $\vec{RT}$ makes an angle of $30^\circ$ (or $150^\circ$) with $\vec{PQ}$.

Because the angle between $\vec{PQ}$ and $\vec{RT}$ is not $0^\circ$ or $180^\circ$, the vectors $\vec{PQ}$ and $\vec{RT}$ are not collinear:
$\vec{PQ} \times (\vec{RS} + \vec{ST}) = \vec{PQ} \times \vec{RT} \neq \vec{0}$
Therefore, Statement-1 is True.

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Conclusion:
Statement-1 is True and Statement-2 is False, which corresponds to option (C).