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., $G(1,1,-2)$

Let vertices of the $\triangle A B C$ are $A\left(x_1, y_1, z_1\right), B\left(x_2, y_2, z_2\right)$ and $C\left(x_3, y_3, z_3\right)$, then the mid-point of $B C(5,7,11)$.

$$\begin{aligned} & 5=\frac{x_2+x_3}{2} \Rightarrow x_2+x_3=10 \quad \text{... (i)}\\ & 7=\frac{y_2+y_3}{2} \Rightarrow y_2+y_3=14 \quad \text{... (ii)}\\ & 11=\frac{z_2+z_3}{2} \Rightarrow z_2+z_3=22\quad \text{... (iii)} \end{aligned}$$

Similarly for the sides AB and AC,

$$\begin{aligned} & 2=\frac{x_1+x_2}{2} \Rightarrow x_1+x_2=4 \quad \text{.... (iv)}\\ & 3=\frac{y_1+y_2}{2} \Rightarrow y_1+y_2=6 \quad \text{.... (v)}\\ & 1=\frac{z_1+z_2}{2} \Rightarrow z_1+z_2=-2 \quad \text{.... (vi)}\\ & 0=\frac{x_1+x_3}{2} \Rightarrow x_1+x_3=0 \quad \text{.... (vii)}\\ & 8=\frac{y_1+y_3}{2} \Rightarrow y_1+y_3=16 \quad \text{.... (viii)}\\ & 5=\frac{z_1+z_3}{2} \Rightarrow z_1+z_3=10\quad \text{.... (ix)} \end{aligned}$$

From Eqs. (i) and (iv),

$x_1+2x_2+x_3=14\quad \text{.... (x)}$

From Eqs. (ii) and (v),

$y_1+2y_2+y_3=20\quad \text{.... (xi)}$

From Eqs. (iii) and (vi),

$z_1+2z_2+z_3=20\quad \text{.... (xii)}$

From Eqs. (vii) and (x),

$$\begin{aligned} 2 x_2 & =14 \Rightarrow x_2=7 \\ x_2 & =7, \text { then } x_3=10-7=3 \\ x_3 & =3 \text {, then } x_1=-3 \\ x_1 & =-3, x_2=7, x_3=3 \end{aligned}$$

From Eqs. (viii) and (xi),

$$\begin{aligned} 2 y_2 & =4 \Rightarrow y_2=2 \\ y_2 & =2, \text { then } y_1=4 \\ y_1 & =4, \text { then } y_3=12 \\ y_1 & =4, y_2=2, y_3=12 \end{aligned}$$

From Eqs. (ix) and (xii),

$$\begin{aligned} &\begin{aligned} 2 z_2 & =10 \Rightarrow z_2=5 \\ z_2 & =5, \text { then } z_1=-7 \\ z_1 & =-7, \text { then } z_3=17 \\ z_1 & =-7, z_2=5, z_3=17 \end{aligned}\\ &\text { So, the vertices are } A(-3,4,-7), B(7,2,5) \text { and } C(3,12,17) \text {. } \end{aligned}$$

$$\begin{aligned} &\text { Let the fourth vertex of the parallelogram } A B C D \text { is } D(x, y, z) \text {. Then, the mid-point of } A C \text { are }\\ &P\left(\frac{1+2}{2}, \frac{2+3}{2}, \frac{3+2}{2}\right) \text { i.e., } P\left(\frac{3}{2}, \frac{5}{2}, \frac{5}{2}\right) \text {. } \end{aligned}$$

$$\begin{aligned} &\text { Now, mid-point of } B D \text {, }\\ &\begin{aligned} & \frac{3}{2}=\frac{-1+x}{2} \Rightarrow x=4 \\ & \frac{5}{2}=\frac{-2+y}{2} \Rightarrow y=7 \\ & \frac{5}{2}=\frac{-1+z}{2} \Rightarrow z=6 \end{aligned} \end{aligned}$$

So, the coordinates of fourth vertex is (4, 7, 6).

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., $G(0,0,0)$.

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