Let A, B, C, D be the points in the plane with position vectors $-2\hat{i} - \hat{j}$, $4\hat{i}$, $3\hat{i} + 3\hat{j}$ and $-3\hat{i} + 2\hat{j}$ respectively, then $\square$ABCD is

The position vectors of the vertices are given as $\vec{A} = -2\hat{i} - \hat{j}$, $\vec{B} = 4\hat{i}$, $\vec{C} = 3\hat{i} + 3\hat{j}$, and $\vec{D} = -3\hat{i} + 2\hat{j}$.

The vectors representing the sides of the quadrilateral are:
$\vec{AB} = \vec{B} - \vec{A} = (4 - (-2))\hat{i} + (0 - (-1))\hat{j} = 6\hat{i} + \hat{j}$
$\vec{BC} = \vec{C} - \vec{B} = (3 - 4)\hat{i} + (3 - 0)\hat{j} = -\hat{i} + 3\hat{j}$
$\vec{CD} = \vec{D} - \vec{C} = (-3 - 3)\hat{i} + (2 - 3)\hat{j} = -6\hat{i} - \hat{j}$
$\vec{DA} = \vec{A} - \vec{D} = (-2 - (-3))\hat{i} + (-1 - 2)\hat{j} = \hat{i} - 3\hat{j}$

Since $\vec{AB} = -\vec{CD}$ and $\vec{BC} = -\vec{DA}$, the opposite sides are equal and parallel. Thus, ABCD is a parallelogram.

The lengths of adjacent sides are:
$|\vec{AB}| = \sqrt{6^2 + 1^2} = \sqrt{37}$
$|\vec{BC}| = \sqrt{(-1)^2 + 3^2} = \sqrt{10}$

Since $|\vec{AB}| \neq |\vec{BC}|$, the adjacent sides are not equal in length, so ABCD is not a rhombus.

The dot product of adjacent side vectors is:
$\vec{AB} \cdot \vec{BC} = (6)(-1) + (1)(3) = -6 + 3 = -3$

Since $\vec{AB} \cdot \vec{BC} \neq 0$, the angle between adjacent sides is not $90^{\circ}$, so ABCD is not a rectangle.

Therefore, ABCD is a parallelogram which is neither a rhombus nor a rectangle.

Answer: a parallelogram which is neither a rhombus nor a reactangle