Students of a school are standing in rows and columns in their playground for a drill practice. $A, B, C$ and $D$ are the positions of four students as shown in the figure. Is it possible to place Jaspal in the drill in such a way that he is equidistant from each of the four students $A, B, C$ and $D$? If so,what should be his position?

(D) Yes,from the figure,we observe that the positions of four students $A, B, C$ and $D$ are $(3, 5), (7, 9), (11, 5)$ and $(7, 1)$ respectively. These are the four vertices of a quadrilateral.
First,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(7 - 3)^2 + (9 - 5)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
$BC = \sqrt{(11 - 7)^2 + (5 - 9)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
$CD = \sqrt{(7 - 11)^2 + (1 - 5)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
$DA = \sqrt{(3 - 7)^2 + (5 - 1)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
Since all sides are equal,it is a rhombus.
Next,we calculate the lengths of the diagonals:
$AC = \sqrt{(11 - 3)^2 + (5 - 5)^2} = \sqrt{8^2 + 0^2} = 8$
$BD = \sqrt{(7 - 7)^2 + (1 - 9)^2} = \sqrt{0^2 + (-8)^2} = 8$
Since the diagonals are also equal $(AC = BD)$,the quadrilateral $ABCD$ is a square.
In a square,the point equidistant from all four vertices is the intersection point of the diagonals,which is the midpoint of either diagonal.
Using the midpoint formula for $AC$:
$P = \left(\frac{3 + 11}{2}, \frac{5 + 5}{2}\right) = \left(\frac{14}{2}, \frac{10}{2}\right) = (7, 5)$
Thus,Jaspal should be placed at position $(7, 5)$.