The plane parallel to $Y Z$-plane is perpendicular to $X$-axis.

Given dimensions are $a=10, b=13$ and $c=8$.

$$\begin{aligned} \therefore \quad \text { Required length } & =\sqrt{a^2+b^2+c^2} \\ & =\sqrt{100+169+64}=\sqrt{333} \end{aligned}$$

Given points are $(a, 2,1)$ and $(1,-1,1)$.

$$\begin{array}{lr} \therefore & \sqrt{(a-1)^2+(2+1)^2+(1-1)^2}=5 \\ \Rightarrow & (a-1)^2+9+0=25 \\ \Rightarrow & a^2-2 a+1+9=25 \\ \Rightarrow & a^2-2 a-15=0 \\ \Rightarrow & a^2-5 a+3 a-15=0 \\ \Rightarrow & a(a-5)+3(a-5)=0 \\ \Rightarrow & (a-5)(a+3)=0 \\ \Rightarrow & a-5=0 \text { or } a+3=0 \\ \therefore & a=+5 \text { or }-3 \end{array}$$

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

$$\begin{aligned} &\text { Since, } D \text { is the mid-point of } A B \text {, then }\\ &\begin{aligned} & \frac{x_1+x_2}{2}=1 \Rightarrow x_1+x_2=2 \quad \text{.... (i)}\\ & \frac{y_1+y_2}{2}=2 \Rightarrow y_1+y_2=4 \quad \text{.... (ii)}\\ & \frac{z_1+z_2}{2}=-3 \Rightarrow z_1+z_2=-6\quad \text{.... (iii)} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { Similarly, } E \text { and } F \text { are the mid-points of sides } B C \text { and } A C \text {, respectively. }\\ &\begin{aligned} & \frac{x_2+x_3}{2}=3 \Rightarrow x_2+x_3=6 \quad \text{.... (iv)}\\ & \frac{y_2+y_3}{2}=0 \Rightarrow y_2+y_3=0 \quad \text{.... (v)}\\ & \frac{z_2+z_3}{2}=1 \Rightarrow z_2+z_3=2 \quad \text{.... (vi)}\\ & \frac{x_1+x_3}{2}=-1 \Rightarrow x_1+x_3=-2 \quad \text{.... (vii)}\\ & \frac{y_1+y_3}{2}=1 \Rightarrow y_1+y_3=2 \quad \text{.... (viii)}\\ & \frac{z_1+z_3}{2}=-4 \Rightarrow z_1+z_3=-8\quad \text{.... (ix)} \end{aligned} \end{aligned}$$

From Eqs. (i) and (iv), $x_1+2 x_2+x_3=8\quad \text{... (x)}$

From Eqs. (ii) and (v), $y_1+2 y_2+y_3=4\quad\text{.... (xi}$

From Eqs. (iii) and (vi), $z_1+2 z_2+z_3=-4\quad \text{.... (xi)}$

From Eqs. (vii) and (x), $2 x_2=10 \Rightarrow x_2=5$

$$\begin{array}{ll} \Rightarrow & x_2=5 \text {, then } x_3=1 \\ \text { If } x_3=1 \text {, then } x_1=-3 \\ \because \quad & x_1=-3, x_2=5, x_3=1 \end{array}$$

$$\begin{aligned} &\text { From Eqs. (viii) and (xi), }\\ &\begin{aligned} 2 y_2 & =2 \Rightarrow y_2=1 \\ \text{If}\quad y_2 & =1, \text { then } y_3=-1 \\ \text{If}\quad y_3 & =-1, \text { then } y_1=3 \\ \because \quad y_1 & =3, y_2=1, y_3=-1 \end{aligned} \end{aligned}$$

From Eqs (ix) and (xii),

$$\begin{aligned} 2 z_2 & =4 \Rightarrow z_2=2 \\ \text{if}\quad z_2 & =2, \text { then } z_3=0 \\ \text{if}\quad z_3 & =0, \text { then } z_1=-8 \\ \because\quad z_1 & =-8, z_2=2, z_3=0 \end{aligned}$$

So, the vertices of $\triangle A B C$ are $A(-3,3,-8), B(5,1,2)$ and $C(1,-1,0)$.

Hence, coordinates of centroid of $\triangle A B C, G\left(\frac{-3+5+1}{3}, \frac{3+1-1}{3}, \frac{-8+2+0}{3}\right)$

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

(i) In XY-plane, Z-coordinates is zero.

(ii) The point $(2,3,4)$ lies in 1 st octant .

(iii) Locus of the points having $x$-coordinate is zero is YZ-plane.

(iv) A line is parallel to $X$-axis if and only if all the points on the line have equal $y$ and $z$-coordinates.

(v) $x=0, y=0$ represent $Z$-axis.

(vi) $z=c$ represent the plane parallel to $X Y$-plane.

(vii) The planes $x=a, y=b$ represent the line parallel to $Z$-axis.

viii) Coordinates of a point are the distances from the origin to the feet of perpendicular from the point on the respective.

(ix) A ball is the solid region in the space enclosed by a sphere.

(x) The region in the plane enclosed by a circle is known as a disc.