Find the centroid of the triangle mid points of whose sides are `(1,2,-3),(3,0,1)and (-1,1,4)`

Explanation

Given that, mid - points of sides are D ( 1,2,-3) E(3, 0, 1) and F ( -1,1,-4)
Let the vertices of the `DeltaABC` are `(A(x_(1).y_(1).z_(1)) and B( x_(2), y_(2), z_(3)) and C(x_(3), y_(3), z_(3))`
then mid point of BC are ( 1,2,-3)
`1=(x_(2)+x_(3))/2Rightarrowx_(2)+x_(3)=2`
`2=(y_(2)+y_(3))/2Rightarrowy_(2)+y_(3)=4`
`-3= (z_(2)+z_(3))/2Rightarrowz_(2)+z_(3)=-6`
Similarly for the sides AB and AC .
`-1= (x_(1)+x_(2))/2Rightarrowx_(1)+x_(2)=-2`
` 1=(y_(1)+y_(2))/2Rightarrowy_(1)+y_(2)=2`
`-4= (z_(1)+z_(2))/2Rightarrowz_(1)+z_(2)=-8`
`3=(x_(1)+y_(3))/2Rightarrowx_(1)+x_(3)=6`
`0= (y_(1)+y_(3))/2Rightarrowy_(1)+y_(3)=0`
`1= (z_(1)+z_(3))/2Rightarrowz_(1)+z_(3)=2`
on adding Eqs, ( i) and (iv) ,we get
`x_(1) + 2x_(2)+x_(3) =0`
on adding Eqs. (ii) and (v) we get
`y_(1)+2y_(2)+y_(3)=6`
On adding Eqs (iii) and (vi) we get
` z_(1)+2z_(2)=z_(3)=-14`
from Eqs. (Eqs. (vii) and (x) .
`2y_(2)= -6 Rightarrowx_(2)=-3`
lf `x_(2)=-3, then x_(3)=5`
lf `x_(3)=5 then x_(1)=1, x_(2)=-3, x_(3)=5`
from Eqs. (xii) and (viii)
`2y_(2)=6 Rightarrow y_(2)=3`
if `y_(2)=3, "then" y_(1)=-1 if y_(1)=-1 "then" y_(3)=1, y_(2)=3,y_(3)=1`
from Eqs. (xii) and (ix) .
`2z_(2)=-16Rightarrowz_(2)=-8`
`z_(2)=-8"then"z_(1)=0`
`z_(1)=0,z_(2)=-8,z_(3)=2`
so, the points are `A (1-10) , B (-3,3,-8)and C (5,1,2)`
centroid of the triangle = `G ((1-3+5)/3,(-1+3+1)/3,(0-8+2)/3)` i,g,(1,1-2)