The mid-points of the sides of a triangle are $(5,7,11)$,$(0,8,5)$,and $(2,3,-1)$. Find its vertices.

(N/A) Let the vertices of the triangle be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.
Given mid-points are $D(5,7,11)$ on $BC$,$E(0,8,5)$ on $AC$,and $F(2,3,-1)$ on $AB$.
Using the mid-point formula:
$1) \frac{x_2+x_3}{2} = 5, \frac{y_2+y_3}{2} = 7, \frac{z_2+z_3}{2} = 11 \Rightarrow x_2+x_3=10, y_2+y_3=14, z_2+z_3=22$
$2) \frac{x_1+x_3}{2} = 0, \frac{y_1+y_3}{2} = 8, \frac{z_1+z_3}{2} = 5 \Rightarrow x_1+x_3=0, y_1+y_3=16, z_1+z_3=10$
$3) \frac{x_1+x_2}{2} = 2, \frac{y_1+y_2}{2} = 3, \frac{z_1+z_2}{2} = -1 \Rightarrow x_1+x_2=4, y_1+y_2=6, z_1+z_2=-2$
Adding all equations for $x$:
$(x_2+x_3) + (x_1+x_3) + (x_1+x_2) = 10+0+4 = 14$ $\Rightarrow 2(x_1+x_2+x_3) = 14$ $\Rightarrow x_1+x_2+x_3 = 7$
$x_1 = (x_1+x_2+x_3) - (x_2+x_3) = 7-10 = -3$
$x_2 = (x_1+x_2+x_3) - (x_1+x_3) = 7-0 = 7$
$x_3 = (x_1+x_2+x_3) - (x_1+x_2) = 7-4 = 3$
Similarly for $y$:
$y_1+y_2+y_3 = (14+16+6)/2 = 18$
$y_1 = 18-14 = 4, y_2 = 18-16 = 2, y_3 = 18-6 = 12$
Similarly for $z$:
$z_1+z_2+z_3 = (22+10-2)/2 = 15$
$z_1 = 15-22 = -7, z_2 = 15-10 = 5, z_3 = 15-(-2) = 17$
Thus,the vertices are $A(-3, 4, -7)$,$B(7, 2, 5)$,and $C(3, 12, 17)$.