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

Let the vertices of a triangle are Ax1,y1,z1,Bx2,y2,z2 and Cx3,y3,z3. Since, D, E and F are the mid-points of AC ,BC and AB∴x1+x22,y1+y22,z1+z22=(0,8,5)⇒x1+x2=0,y1+y2=16,z1+z2=10-----i x2+x32,y2+y32,z2+z32=(2,3,−1)⇒x2+x3=4,y2+y3=6,z2+z3=−2----ii and x1+x32,y1+y32,z1+z32=(5,7,11)⇒x1+x3=10,y1+y3=14,z1+z3=22----iiiOn adding Eqs. (i), (ii) and (iir), we get2x1+x2+x3=14,2y1+y2+y3=36,2z1+z2+z3=30⇒x1+x2+x3=7,y1+y2+y3=18,z1+z2+z3=15… (iv) On solving Eqs. (i), (ii), (iil) and (iv)' we getx3=7,x1=3,x2=−3y3=2,y1=12,y2=4and  z3=5,z1=17,z2=−7Hence, vertices of a triangle are (7, 2,5), (3, 2, 17), (-3, 4, -7)