Step-by-Step Solution:

1. Identify the position vectors of the vertices. Let the origin be $O$. The position vectors of the points $P, Q, R,$ and $S$ are given as:

   • $\vec{OP} = \vec{p} = -2\widehat i - \widehat j$
   • $\vec{OQ} = \vec{q} = 4\widehat i$
   • $\vec{OR} = \vec{r} = 3\widehat i + 3\widehat j$
   • $\vec{OS} = \vec{s} = -3\widehat i + 2\widehat j$

2. Calculate the vectors representing the sides of the quadrilateral. We find the vectors for each side of the quadrilateral $PQRS$:

   • $\vec{PQ} = \vec{q} - \vec{p} = (4\widehat i) - (-2\widehat i - \widehat j) = 6\widehat i + \widehat j$
   • $\vec{QR} = \vec{r} - \vec{q} = (3\widehat i + 3\widehat j) - (4\widehat i) = -\widehat i + 3\widehat j$
   • $\vec{RS} = \vec{s} - \vec{r} = (-3\widehat i + 2\widehat j) - (3\widehat i + 3\widehat j) = -6\widehat i - \widehat j$
   • $\vec{SP} = \vec{p} - \vec{s} = (-2\widehat i - \widehat j) - (-3\widehat i + 2\widehat j) = \widehat i - 3\widehat j$

3. Check if the quadrilateral is a parallelogram. A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel, which means their corresponding vectors are equal. Let's compare $\vec{PQ}$ and $\vec{SR}$. Note that $\vec{SR} = -\vec{RS}$.

   • $\vec{PQ} = 6\widehat i + \widehat j$
   • $\vec{SR} = -\vec{RS} = -(-6\widehat i - \widehat j) = 6\widehat i + \widehat j$ Since $\vec{PQ} = \vec{SR}$, the sides $PQ$ and $SR$ are equal in length and parallel. Therefore, the quadrilateral $PQRS$ is a parallelogram. Alternatively, we can check the other pair of opposite sides:
   • $\vec{PS} = -\vec{SP} = -(\widehat i - 3\widehat j) = -\widehat i + 3\widehat j$
   • $\vec{QR} = -\widehat i + 3\widehat j$ Since $\vec{PS} = \vec{QR}$, this also confirms that $PQRS$ is a parallelogram.

4. Check if the parallelogram is a rhombus. A rhombus is a parallelogram with all sides of equal length. Let's find the magnitudes (lengths) of the adjacent sides $\vec{PQ}$ and $\vec{QR}$.

   • $|\vec{PQ}| = \sqrt{6^2 + 1^2} = \sqrt{36 + 1} = \sqrt{37}$
   • $|\vec{QR}| = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$ Since $|\vec{PQ}| \neq |\vec{QR}|$, the adjacent sides are not equal. Thus, the parallelogram is not a rhombus. This also implies it cannot be a square.

5. Check if the parallelogram is a rectangle. A rectangle is a parallelogram with perpendicular adjacent sides. We can check this by taking the dot product of the vectors of two adjacent sides. If the dot product is zero, the sides are perpendicular.

   • $\vec{PQ} \cdot \vec{QR} = (6\widehat i + \widehat j) \cdot (-\widehat i + 3\widehat j)$
   • $= (6)(-1) + (1)(3) = -6 + 3 = -3$ Since $\vec{PQ} \cdot \vec{QR} \neq 0$, the adjacent sides are not perpendicular. Therefore, the parallelogram is not a rectangle. This also implies it cannot be a square.

   Alternatively, we can check if the diagonals are equal.

   • $\vec{PR} = \vec{r} - \vec{p} = (3\widehat i + 3\widehat j) - (-2\widehat i - \widehat j) = 5\widehat i + 4\widehat j$
   • $|\vec{PR}| = \sqrt{5^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41}$
   • $\vec{QS} = \vec{s} - \vec{q} = (-3\widehat i + 2\widehat j) - (4\widehat i) = -7\widehat i + 2\widehat j$
   • $|\vec{QS}| = \sqrt{(-7)^2 + 2^2} = \sqrt{49 + 4} = \sqrt{53}$ Since $|\vec{PR}| \neq |\vec{QS}|$, the diagonals are not equal, which confirms it is not a rectangle.

6. Conclusion. The quadrilateral $PQRS$ is a parallelogram, but it is neither a rhombus (unequal adjacent sides) nor a rectangle (adjacent sides not perpendicular).

   Therefore, the correct description is a parallelogram, which is neither a rhombus nor a rectangle.