The coordinates of the mid points of sides AB, BC and CA of ` A B C` are `D(1,2,-3),\ E(3,0,1)a n d\ F(-1,1,-4)` respectively. Write the coordinates of its centroid.

Explanation

Let the vertices of `DeltaABC` is `(x_(1),y_(1),z_(1)) B(x_(2),y_(2),z_(2))and c(x_(3),y_(3),z_(3))`

Since, D is the mid-point of AB then
`(x_(1)+x_(2))/2=1Rightarrowx_(1)+x_(2)=2`
`(y_(1)+y_(2))/2=2Rightarrowy_(1)+y_(2)=4`
`(z_(1)+z_(2))/2=-3 Rightarrowz_(1)+z_(2)=-6`
Similarly E and F are the mid - points of sides BC and AC , respectively
`(x_(2)+x_(3))/2=3Rightarrowx_(2)+x_(3)=6`
`(y_(2)+y_(3))/2=0Rightarrowy_(2)+y_(3)=0`
`(z_(2)+z_(3))/2=1Rightarrowz_(2)+z_(3)=2`
`(x_(1)+x_(2))/2=-1Rightarrowx_(1)+x_(3)=-2`
`(y_(1)+y_(3))/2=1Rightarrow y_(1)+y_(3)=2`
`(z_(1)+z_(3))/2=-4Rightarrowz_(1)+z_(3)=-8`
from Eqs. (i) and (iv)
`x_(1)+2x_(2)+x_(3)=2`
from Eqs. (ii) and (v),
`y_(1)+2y_(2)+y_(3)=4`
from Eqs. (iii) and (vi) ,
`z_(1) + 2z_(3)+z_(3)=-4`
from Eqs. (vii) and (x)
`2x_(2) =10Rightarrow x_(2)=5`
`x_(2)=5, " then " x_(3)=1`
if `x_(3)=1, "then" x_(1)=-3`
` x_(1)=-3, x_(2)=5, x_(3)=1`
from Eqs. (ix) and (xi)
`2y_(1)=2Rightarrowy_(2)=1`
`y_(2)=1,"then" y_(3)=-1`
`y_(3)=-1, "then" y_(1)=3`
`y_(1)=3, y_(2)=1,y_(3)=-1`
from Eqs (ix) and (xii)
`2z_(2)=4Rightarrowz_(2)=2`
`z_(2)=2,"then" z_(3)=0`
`z_(3)=0, "then" z_(1)=-8`
`z_(1)=-8, z_(2)=2,z_(3)=0`
So the vertices of `DeltaABC` are A (-3,3,-8) (5,1,2) and C ( 1,-1,0)
Hence, coordinates of centroid of `DeltaABC, G((-3+5+1)/3,(3+1-1)/3,(-8+2+0)/3)`
G(1,1,-2)