(1,0,-3) >"> ({-1},{5/3},{-3}) Solution: Let vertices of \triangle A B C are A(x_{1}, y_{1}, z_{1}), B(x_{2}, y_{2}, z_{2}) and C(x_{3}, y_{3}, z_{3}) ∵ D, E and F are mid-points of A B, B C and A C respectively. ( {x_{1}+x_{2}}/{2}, {y_{1}+y_{2}}/{2}, {z_{1}+z_{2}}/{2})=(1,2,-3) coordinate of D. ( {x_{2}+x_{3}}/{2}, {y_{2}+y_{3}}/{2}, {z_{2}+z_{3}}/{2})=(3,0,1) coordinate of E ( {x_{1}+x_{3}}/{2}, {y_{1}+y_{3}}/{2}, {z_{1}+z_{3}}/{2})=(-1,1,-4) coordinate of F Adding all three, (x_{1}+x_{2}+x_{3}, y_{1}+y_{2}+y_{3}, z_{1}+z_{2}+z_{3})=(3,3,-6) Also, (x_{2}+x_{3}, y_{2}+y_{3}, z_{2}+z_{3})=(6,0,2) ∴ A(x_{1}, y_{1}, z_{1})=(-3,3,-8) Centroid of \triangle A D F where A(-3,3,-8), D(1,2,-3) and F(-1,1,-4) is (\frac{-3+1-1}{3}, \frac{3+2+1}{3}, \frac{-8-3-4}{3})=(-1,2,-5)" >
