Answer :

The midpoints of the sides of a triangle are (1, 5, -1), (0, 4, -2) and (2, 3, 4).


Let its vertices be A(x1,y1,z1), B(x2,y2,z2), C(x3,y3,z3) .


The mid point of AB is (1,5,-1), therefore



x1 +x2 = 2………..eq.1



y1 + y2 = 10……eq.2



z1 + z2 = -2……..eq.3


Mid point of AC is (2,3,4), therefore



x1 +x3 = 4………..eq.4



y1 + y3 = 6……eq.5



z1 + z3 = 8……..eq.6


Mid point of BC is (0,4,-2), therefore



x2 +x3 = 0………..eq.7



y3 + y2 = 8……eq.8



z3 + z2 = -4……..eq.9


now, adding the equations 1,4 and 7, and divide it by two we get,


x1 + x2 + x3 = 3


now subtracting 1, 4, 7 individually, we get


x1 = 3, x2 = -1 and x3 = 1


now, adding the equations 2,5 and 8, and divide it by two we get,


y1 + y2 + y3 = 12


now subtracting 1, 4, 7 individually, we get


y1 = 4, y2 = 6 and y3 = 2


now, adding the equations 3,6 and 9, and divide it by two we get,


z1 + z2 + z3 = 1


now subtracting 1, 4, 7 individually, we get


z1 = 5, z2 = -7 and z3 = 3


therefore, the coordinates are A(3,4,5), B(-1,6,-7) and C(1,2,3).


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