Given: Mid-point of the sides of a triangle AB, BC, CA are $D\left(1,2,-3\right), E\left(3,0,1\right), F\left(-1,1,-4\right)$

Let the vertices of $∆ABC A\left(x_1,y_2,z_1\right), B\left(x_2,y_2,z_2\right), C\left(x_3,y_3,z_3\right)$and $D$ is the mid-point of $AB$ then

Co-ordinate of point $D$ is

$\frac{x_1+x_2}{2}=1 \Rightarrow x_1+x_2=2 ...\left(1\right)$

$\frac{y_1+y_2}{2}=2 \Rightarrow y_1+y_2=1 ...\left(2\right),$

$\frac{z_1+z_2}{2}=-3 \Rightarrow z_1+z_2=-6 ...\left(3\right)$

Similarly, E and F are the mid-point of side BC and AC.

Co-ordinate of point $E$is

$\Rightarrow \frac{x_2+x_3}{2}=3 \Rightarrow x_2+x_3=6 ...\left(4\right)$

$\Rightarrow \frac{y_2+y_3}{2}=0 \Rightarrow y_2+y_3=0 ...\left(5\right)$

$\Rightarrow \frac{z_2+z_3}{2}=1 \Rightarrow z_2+z_3=2 ...\left(6\right)$

And Co-ordinate of point $F$is

$\frac{x_1+x_3}{2}=-1 \Rightarrow x_1+x_3=-2 ...\left(7\right)$

$\frac{y_1+y_3}{2}=1 \Rightarrow y_1+y_3=2 ...\left(8\right)$

$\frac{z_1+z_3}{2}=-4 \Rightarrow z_1+z_3=-8 ...\left(9\right)$

From equations $\left(1\right),\left(4\right),\left(7\right) \Rightarrow x_1=-3, x_2=5, x_3=1$

From equations $\left(2\right),\left(5\right),\left(8\right) \Rightarrow y_1=3, y_2=1, y_3=-1$

From equations $\left(3\right),\left(6\right),\left(9\right) \Rightarrow z_1=-8, z_2=2, z_3=0$

So, $A(-3,3,-8),B(5,1,2) \& C(1,-1,0)$

Hence, the vertices of $∆ADF$ are $A\left(-3,3,-8\right), \mathrm{D}\left(1, 2, -3\right) \mathrm{and} F\left(-1, 1, -4\right)$

So, co-ordinate of centroid of $∆ADF$are

$G\left(\frac{-3+1-1}{3},\frac{3+2+1}{3},\frac{-8-3-4}{3}\right)=G\left(-1,2,-5\right)$

Hence, centroid $G\left(-1,2,-5\right)$