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

Explanation

Let vertices of the `DeltaABC` are `A (x_(1),y_(1),z_(1)), B(x_(2),y_(2),z_(3))and C(x_(3),y_(3),z_(3))` then the mid-point of BC ( 5,7,11)
`5=(x_(2)+x_(3))/2Rightarrowx_(2)+x_(3)=10`
`7=(y_2+y_(3))/2 Rightarrowy_(2)+y_(3)=14`
`11=(z_(2)+z_(3))/2Rightarrow z_(2)+z_(3)=22`
Similarly for the sides AB and AC.
`2=(x_(1)+x_(2))/2 Rightarrow x_(1)=x_(2)=4`
`3=(y_(1)+y_(2))/2 Rightarrow y_(1)=y_(2)=6`
`-1=(z_(1)+z_(2))/2 Rightarrow z_(1)=z_(2)=-2`
`0=(x_(1)+x_(3))/2Rightarrowx_(1)+x_(3)=0`
`8=(y_(1)+y_(3))/2Rightarrowy_(1)+y_(3)=16`
`5=(z_(1)+z_(3))/2Rightarrowz_(1)+z_(3)=10`
Form Eqs. (i) and (iv)
`x_(1)+2x_(2)+x_(2)=14`
from Eqs. (ii) and (vi)
`y_(1)+2y_(2)+y_(3)=20`
from Eqs. (iii) and (vi)
`z_(1)+2z_(2)+z_(3)+20`
from Eqs. (vii) and (x)
`2x_(2)=14Rightarrowx_(2)=7`
`x_(2)=7, "then "x_(3)=10-7=3`
`x_(3)=3, "then"x_(1)=-3`
`x_(1)=-3,x_(2)=7,x_(3)=12`
from Eqs. (viii) and (xi),
`2y_(2)=4Rightarrowy_(2)=5`
`y_(2)=2,"then"y_(1)=4`
`y(1)=4, "then"y_(3)=12`
`y_(1)=4,y_(2)=2,y_(3)=12`
From Eqs. (ix) and (xii)
`2 z_(2)=10Rightarrowz_(2)=5`
`z_(2)=5, " then" z_(1)=-7`
`z_(1)=-7,"then" z_(3)=17`
`z_(1)=-7,z_(2)=5,z_(3)=17`
So, the vertices are A (-3,4,-7) B (7,2,5) and C ( 3,12,17)