Given points are $A(2,3,4), B(-1,2,-3)$ and $C(-4,1,-10)$.

$$\begin{aligned} & \therefore \quad A B=\sqrt{(2+1)^2+(3-2)^2+(4+3)^2} \\ & =\sqrt{9+1+49}=\sqrt{59} \\ & B C=\sqrt{(-1+4)^2+(2-1)^2+(-3+10)^2} \\ & =\sqrt{9+1+49}=\sqrt{59} \\ & A C=\sqrt{(2+4)^2+(3-1)^2+(4+10)^2} \\ & =\sqrt{36+4+196} \\ & =\sqrt{236}=2 \sqrt{59} \\ & \text { Now, } \\ & A B+B C=\sqrt{59}+\sqrt{59}=2 \sqrt{59} \\ & \because \quad A B+B C=A C \end{aligned}$$

Hence, the points $A, B$ and $C$ are collinear.

Now, $$A C: B C=2 \sqrt{59}: \sqrt{59}=2: 1$$

So, $C$ divide $A B$ in $2: 1$ externally.

Let the vertices of $\triangle A B C$ are $A\left(x_1, y_1, z_1\right), B\left(x_2, y_2, z_2\right)$ and $C\left(x_3, y_3, z_3\right)$.

Since, the mid-point of side $B C$ is $D(1,5,-1)$.

Then,

$$\begin{aligned} & \frac{x_2+x_3}{2}=1 \Rightarrow x_2+x_3=2 \quad \text{.... (i)}\\ & \frac{y_2+y_3}{2}=5 \Rightarrow y_2+y_3=10 \quad \text{.... (ii)}\\ & \frac{z_2+z_3}{2}=-1 \Rightarrow z_2+z_3=-2\quad \text{.... (iii)} \end{aligned}$$

Similarly, the mid-points of $A B$ and $A C$ are $F(2,3,4)$ and $E(0,4,-2)$,

$$\begin{aligned} & \frac{x_1+x_2}{2}=2 \Rightarrow x_1+x_2=4 \quad \text{.... (iv)}\\ & \frac{y_1+y_2}{2}=3 \Rightarrow y_1+y_2=6 \quad \text{.... (v)}\\ \text{and}\quad & \frac{z_1+z_2}{2}=4 \Rightarrow z_1+z_2=8\quad \text{.... (vi)} \end{aligned}$$

Now,

$$\begin{gathered} \frac{x_1+x_3}{2}=0 \Rightarrow x_1+x_3=0 \quad \text{.... (vii)}\\ \frac{y_1+y_3}{2}=4 \Rightarrow y_1+y_3=8 \quad \text{.... (viii)}\\ \frac{z_1+z_3}{2}=-2 \Rightarrow z_1+z_3=-4\quad \text{.... (ix)} \end{gathered}$$

$$\begin{aligned} &\text { From Eqs. (i) and (iv), }\\ &x_1+2 x_2+x_3=6\quad \text{.... (x)} \end{aligned}$$

$$\begin{aligned} &\text { From Eqs. (ii) and (v), }\\ &y_1+2 y_2+y_3=16\quad \text{.... (xi)} \end{aligned}$$

$$\begin{aligned} &\text { From Eqs. (iii) and (vi), }\\ &z_1+2 z_2+z_3=6\quad \text{.... (xii)} \end{aligned}$$

$$\begin{aligned} &\text { From Eqs. (vii) and (x), }\\ &\begin{aligned} 2 x_2 & =6 \Rightarrow x_2=3 \\ x_2 & =3, \text { then } x_3=-1 \\ x_3 & =-1 \end{aligned} \end{aligned}$$

Then, $$x_1=1 \Rightarrow x_1=1, x_2=3, x_2=-1$$

From Eqs. (viii) and (xi),

$$\begin{aligned} 2 y_2 & =8 \Rightarrow y_2=4 \\ y_2 & =4 \end{aligned}$$

Then,

$$\begin{aligned} & y_1=2 \\ & y_1=2 \\ \text{Then,}\quad & y_3=6 \\ \Rightarrow \quad & y_1=2, y_2=4, y_3=6 \end{aligned}$$

From Eqs. (ix) and (xii),

$$\begin{aligned} 2 z_2 & =10 \Rightarrow z_2=5 \\ z_2 & =5 \\ \text{Then,}\quad z_1 & =3 \\ z_1 & =3 \\ \text{Then,}\quad z_3 & =-7 \\ \Rightarrow \quad z_1 & =3, z_2=5, z_3=-7 \end{aligned}$$

$\Rightarrow \quad z_1=3, z_2=5, z_3=-7$

So, the vertices of the triangle $A(1,2,3), B(3,4,5)$ and $C(-1,6,-7)$.

Hence, centroid of the triangle $G\left(\frac{1+3-1}{3}, \frac{2+4+6}{3}, \frac{3+5-7}{3}\right)$ i.e., $G(1,4,1 / 3)$.

Given points are $A(0,-1,-7), B(2,1,-9)$ and $C(6,5,-13)$

$$\begin{aligned} & A B=\sqrt{(0-2)^2+(-1-1)^2+(-7+9)^2}=\sqrt{4+4+4}=2 \sqrt{3} \\ & B C=\sqrt{(2-6)^2+(1-5)^2+(-9+13)^2}=\sqrt{16+16+16}=4 \sqrt{3} \\ & A C=\sqrt{(0-6)^2+(-1-5)^2+(-7+13)^2}=\sqrt{36+36+36}=6 \sqrt{3} \end{aligned}$$

$$\begin{aligned} &\begin{array}{ll} \because & A B+B C=2 \sqrt{3}+4 \sqrt{3}=6 \sqrt{3} \\ \text { So, } & A B+B C=A C \end{array}\\ &\text { Hence, the points } A, B \text { and } C \text { are collinear. } \end{aligned}$$

$$\begin{aligned} &A B: A C=2 \sqrt{3}: 6 \sqrt{3}=1: 3\\ &\text { So, point } A \text { divide } B \text { and } C \text { in 1: 3 externally. } \end{aligned}$$

The coordinates of the cube which edge is 2 units, are $(2,0,0),(2,2,0),(0,2,0)$, $(0,2,2),(0,0,2),(2,0,2),(0,0,0)$ and $(2,2,2)$.