The points $(3,1)$,$(5,6)$,and $(-3,2)$ are the midpoints of the sides of a triangle. Find the vertices of the triangle.

(N/A) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.
Let the midpoints be $D(3,1)$,$E(5,6)$,and $F(-3,2)$.
Using the midpoint formula,we have:
$(x_1+x_2)/2 = 3, (y_1+y_2)/2 = 1 \implies x_1+x_2 = 6, y_1+y_2 = 2$ (Eq. $1$)
$(x_2+x_3)/2 = 5, (y_2+y_3)/2 = 6 \implies x_2+x_3 = 10, y_2+y_3 = 12$ (Eq. $2$)
$(x_3+x_1)/2 = -3, (y_3+y_1)/2 = 2 \implies x_3+x_1 = -6, y_3+y_1 = 4$ (Eq. $3$)
Adding all equations: $2(x_1+x_2+x_3) = 6+10-6 = 10 \implies x_1+x_2+x_3 = 5$.
Similarly,$y_1+y_2+y_3 = (2+12+4)/2 = 9$.
Subtracting Eq. $2$ from the sum: $x_1 = 5-10 = -5$ and $y_1 = 9-12 = -3$.
Subtracting Eq. $3$ from the sum: $x_2 = 5-(-6) = 11$ and $y_2 = 9-4 = 5$.
Subtracting Eq. $1$ from the sum: $x_3 = 5-6 = -1$ and $y_3 = 9-2 = 7$.
The vertices are $(-5,-3), (11,5), (-1,7)$.