The mid-points $D, E, F$ of the sides of a triangle $ABC$ are $(3, 4), (8, 9),$ and $(6, 7)$ respectively. Find the coordinates of 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 $C(x_3, y_3).$ The mid-points are given as $D(3, 4)$ on $AB,$ $E(8, 9)$ on $BC,$ and $F(6, 7)$ on $AC.$
Using the mid-point formula:
$1) \frac{x_1 + x_2}{2} = 3, \frac{y_1 + y_2}{2} = 4 \implies x_1 + x_2 = 6, y_1 + y_2 = 8$
$2) \frac{x_2 + x_3}{2} = 8, \frac{y_2 + y_3}{2} = 9 \implies x_2 + x_3 = 16, y_2 + y_3 = 18$
$3) \frac{x_1 + x_3}{2} = 6, \frac{y_1 + y_3}{2} = 7 \implies x_1 + x_3 = 12, y_1 + y_3 = 14$
Adding these equations:
$(x_1 + x_2) + (x_2 + x_3) + (x_1 + x_3) = 6 + 16 + 12 = 34 \implies 2(x_1 + x_2 + x_3) = 34 \implies x_1 + x_2 + x_3 = 17$
$(y_1 + y_2) + (y_2 + y_3) + (y_1 + y_3) = 8 + 18 + 14 = 40 \implies 2(y_1 + y_2 + y_3) = 40 \implies y_1 + y_2 + y_3 = 20$
Solving for individual coordinates:
$x_1 = (x_1 + x_2 + x_3) - (x_2 + x_3) = 17 - 16 = 1$
$x_2 = (x_1 + x_2 + x_3) - (x_1 + x_3) = 17 - 12 = 5$
$x_3 = (x_1 + x_2 + x_3) - (x_1 + x_2) = 17 - 6 = 11$
$y_1 = (y_1 + y_2 + y_3) - (y_2 + y_3) = 20 - 18 = 2$
$y_2 = (y_1 + y_2 + y_3) - (y_1 + y_3) = 20 - 14 = 6$
$y_3 = (y_1 + y_2 + y_3) - (y_1 + y_2) = 20 - 8 = 12$
Thus,the vertices are $A(1, 2), B(5, 6),$ and $C(11, 12).$