To determine the type of quadrilateral PQRS is, we need to calculate the properties such as the lengths of the sides and the slopes of the lines that form the sides of the quadrilateral. Let's go through the steps:

1. Calculate the lengths of the sides:

   Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$, we find the lengths of each side.

   • $PQ = \sqrt{(5-0)^2 + (6-3)^2} = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34}$
   • $QR = \sqrt{(2-5)^2 + (11-6)^2} = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$
   • $RS = \sqrt{(2+3)^2 + (11-8)^2} = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34}$
   • $SP = \sqrt{(0+3)^2 + (3-8)^2} = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$

2. Calculate the slopes of the sides:

   Using the slope formula $m = \frac{y_2-y_1}{x_2-x_1}$, we find the slopes of each side.

   • Slope of $PQ = \frac{6-3}{5-0} = \frac{3}{5}$
   • Slope of $QR = \frac{11-6}{2-5} = \frac{5}{-3} = -\frac{5}{3}$
   • Slope of $RS = \frac{11-8}{2+3} = \frac{3}{5}$
   • Slope of $SP = \frac{3-8}{0+3} = \frac{-5}{3} = -\frac{5}{3}$

3. Analyze the results:

   The side lengths are all equal ($\sqrt{34}$), indicating that PQRS is a rhombus since all four sides are equal in length.

   Furthermore, opposite sides have the same slope ($PQ$ is parallel to $RS$ with slope $\frac{3}{5}$, and $QR$ is parallel to $SP$ with slope $-\frac{5}{3}$), and adjacent sides are perpendicular since the product of their slopes is $-1$ (i.e., $\frac{3}{5} \times -\frac{5}{3} = -1$).

Thus, PQRS is a square because it is a rhombus with consecutive sides perpendicular to each other.