Let the $P\left(x_1, y_1, z_1\right)$ and $Q\left(x_2, y_2, z_2\right)$ trisect line segment $A B$.

Since, the point $P$ divided line $A B$ in $1: 2$ internally, then

$$\begin{aligned} & x_1=\frac{2 \times 2+1 \times 5}{1+2}=\frac{9}{3}=3 \\ & y_1=\frac{2 \times 1+1 \times(-8)}{3}=\frac{-6}{3}=-2 \\ & z_1=\frac{2 \times(-3)+1 \times 3}{3}=\frac{-6+3}{3}=\frac{-3}{3}=-1 \end{aligned}$$

Since, the point $Q$ divide the line segment $A B$ in $2: 1$, then

$$\begin{aligned} & x_2=\frac{1 \times 2+2 \times 5}{3}=4 \\ & y_2=\frac{1 \times 1+(-8 \times 2)}{3}=-5 \\ & z_2=\frac{1 \times(-3)+2 \times 3}{3}=-1 \end{aligned}$$

So, the coordinates of $P$ are $(3,-2,-1)$ and the coordinates of $Q$ are $(4,-5,1)$.

Given that origin is the centroid of the $\triangle A B C$ i.e., $G(0,0,0)$.

$$\begin{array}{ll} \because \quad & 0=\frac{a-2+4}{3} \Rightarrow a=-2 \\ & 0=\frac{1+b+7}{3} \Rightarrow b=-8 \\ & 0=\frac{3-5+c}{3} \Rightarrow c=+2 \\ \therefore \quad & a=-2, b=-8 \text { and } c=2 \end{array}$$

Let the coordinates of $D$ are $(x, y, z)$.

$$\begin{aligned} A B & =\sqrt{9+16+144}=\sqrt{169}=13 \\ A C & =\sqrt{0+25+144}=\sqrt{169}=\sqrt{13} \\ \Rightarrow \quad \frac{A B}{A C} & =\frac{13}{13} \Rightarrow A B=A C \\ \frac{B D}{D C} & =\frac{1}{1} \Rightarrow B D=D C \end{aligned}$$

Since, $D$ is divide the line $B C$ in two equal parts. So, $D$ is the mid-point of $B C$.

$$\begin{array}{ll} \therefore & x=\frac{5+2}{2}=7 / 2 \\ \Rightarrow & y=\frac{6+7}{2}=13 / 2 \\ \Rightarrow & z=\frac{9+9}{2}=9 \end{array}$$

So, the coordinates of $D$ are $\left(\frac{7}{2}, \frac{13}{2}, 9\right)$.

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., $G(1,4,1 / 3)$.