$$\begin{aligned} &\text { Given points, } A(1,-1,3), B(2,-4,5) \text { and } C(5,-13,11) \text {. }\\ &\begin{aligned} A B & =\sqrt{(1-2)^2+(-1+4)^2+(3-5)^2} \\ & =\sqrt{1+9+4}=\sqrt{14} \\ B C & =\sqrt{(2-5)^2+(-4+13)^2+(5-11)^2} \\ & =\sqrt{9+81+36}=\sqrt{126} \\ A C & =\sqrt{(1-5)^2+(-1+13)^2+(3-11)^2} \\ & =\sqrt{16+144+64}=\sqrt{224} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\begin{array}{ll} \because & A B+B C=A C \\ \Rightarrow & \sqrt{14}+\sqrt{126}=\sqrt{224} \\ \Rightarrow & \sqrt{14}+3 \sqrt{14}=4 \sqrt{14} \end{array}\\ &\text { So, the points } A, B \text { and } C \text { are collinear. } \end{aligned}$$

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

$$\begin{aligned} \text{Mid-points of diagonal }AC, \quad & x=\frac{x_1+x_2}{2}, y=\frac{y_1+y_2}{2}, z=\frac{z_1+z_2}{2} \\ \text{and}\quad & x=\frac{6-2}{2}=2, y=\frac{-2+2}{2}=0, z=\frac{4+4}{2}=4 \end{aligned}$$

Since, the mid-point of $A C$ are $(2,0,4)$.

Now, mid-point of $B D, 2=\frac{x+2}{2} \Rightarrow x=2$

$$\begin{array}{ll} \Rightarrow & 0=\frac{y+4}{2} \Rightarrow y=-4 \\ \Rightarrow & 4=\frac{z-8}{2} \Rightarrow z=16 \end{array}$$

So, the coordinates of fourth vertex $D$ is $(2,-4,16)$.

Given that, the vertices of the $\triangle A B C$ are $A(0,4,1), B(2,3,-1)$ and $C(4,5,0)$.

Now,

$$\begin{aligned} A B & =\sqrt{(0-2)^2+(4-3)^3+(1+1)^2} \\ & =\sqrt{4+1+4}=3 \\ B C & =\sqrt{(2-4)^2+(3-5)^2+(-1-0)^2} \\ & =\sqrt{4+4+1}=3 \\ A C & =\sqrt{(0-4)^2+(4-5)^2+(1-0)^2} \\ & =\sqrt{16+1+1}=\sqrt{18} \end{aligned}$$

$$\begin{aligned} & \because \quad A C^2=A B^2+B C^2 \\ & \Rightarrow \quad 18=9+9 \end{aligned}$$

Hence, vertices $\triangle A B C$ is a right angled triangle.

Let third vertex of $\Delta ABC$ i.e., is $A(x,y,z)$.

Given that, the coordinate of centroid $G$ are $(0,0,0)$.

$$\begin{aligned} \because \quad 0 & =\frac{x+2+0}{3} \Rightarrow x=-2 \\ 0 & =\frac{y+4-2}{3} \Rightarrow y=-2 \\ 0 & =\frac{z+6-5}{2} \Rightarrow z=-1 \end{aligned}$$

Hence, the third vertex of triangle is $(-2,-2,-1)$.

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

Then, mid-point of $B C$ are $(1,2,-3)$.

$$\begin{aligned} \therefore \quad 1 & =\frac{x_2+x_3}{2} \Rightarrow x_2+x_3=2 \quad \text{... (i)}\\ 2 & =\frac{y_2+y_3}{2} \Rightarrow y_2+y_3=4 \quad \text{... (ii)}\\ -3 & =\frac{z_2+z_3}{2} \Rightarrow z_2+z_3=-6 \quad \text{... (iii)} \end{aligned}$$

Similarly for the sides AB and AC,

$$\begin{array}{ll} \Rightarrow & -1=\frac{x_1+x_2}{2} \Rightarrow x_1+x_2=-2 \quad \text{... (iv)}\\ \Rightarrow & 1=\frac{y_1+y_2}{2} \Rightarrow y_1+y_2=2 \quad \text{... (v)}\\ \Rightarrow & -4=\frac{z_1+z_2}{2} \Rightarrow z_1+z_2=-8 \quad \text{... (vi)}\\ \Rightarrow & 3=\frac{x_1+x_3}{2} \Rightarrow x_1+x_3=6 \quad \text{... (vii)}\\ \Rightarrow & 0=\frac{y_1+y_3}{2} \Rightarrow y_1+y_3=0 \quad \text{... (viii)}\\ \Rightarrow & 1=\frac{z_1+z_3}{2} \Rightarrow z_1+z_3=2 \quad \text{... (ix)} \end{array}$$

On adding Eqs. (i) and (iv), we get

$$x_1+2 x_2+x_3=0\quad \text{.... (x)}$$

On adding Eqs. (ii) and (v), we get

$$y_1+2 y_2+y_3=6\quad \text{.... (xi)}$$

On adding Eqs. (iii) and (vi), we get

$$z_1+2 z_2+z_3=-14\quad \text{.... (xii)}$$

From Eqs. (vii) and (x),

$$2 x_2=-6 \Rightarrow x_2=-3$$

If $x_2=-3$, then $x_3=5$

If $x_3=5$, then $x_1=1, x_2=-3, x_3=5$

From Eqs. (xi) and (viii),

$$2 y_2=6 \Rightarrow y_2=3$$

If $y_2=3$, then $y_1=-1 \quad$ If $y_1=-1$, then $y_3=1, \quad y_2=3, y_3=1$ From Eqs. (xii) and (ix),

$$\begin{aligned} 2 z_2 & =-16 \Rightarrow z_2=-8 \\ z_2 & =-8, \text { then } z_1=0 \\ z_1 & =0, \text { then } z_3=2 \\ z_1 & =0, z_2=-8, z_3=2 \end{aligned}$$

So, the points are $A(1-1,0), B(-3,3,-8)$ and $C(5,1,2)$.

$\therefore$ Centroid of the triangle $=G\left(\frac{1-3+5}{3}, \frac{-1+3+1}{3}, \frac{0-8+2}{3}\right)$ i.e., $G(1,1,-2)$