$$ \begin{aligned} &\text { Given points, } A(2,0,0) \text { and } B(-3,0,0)\\ &A B=\sqrt{(2+3)^2+0^2+0^2}=5 \end{aligned}$$

Distance from origin to the point $(6,6,7)$

$$\begin{aligned} & =\sqrt{(0-6)^2+(0-6)^2+(0-7)^2} \quad\left[\because d=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2+\left(z_1-z_2\right)^2}\right] \\ & =\sqrt{36+36+49} \\ & =\sqrt{121}=11 \end{aligned}$$

Given that, $x^2+y^2=1$

$\therefore$ Distance of the point $\left(x, y, \sqrt{1-x^2-y^2}\right)$ from origin is given as

$$\begin{aligned} d & =\left|\sqrt{x^2+y^2+\left(\sqrt{1-x^2-y^2}\right)^2}\right| \\ & =\left|\sqrt{x^2+y^2+1-x^2-y^2}\right|=1 \end{aligned}$$

Hence proved.

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

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

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 $(2,-4,16)$.