Explanation
$$\textbf{Step -1: Find the distance between the given points.}$$
$$\text{Let }A(5,0,2),B(2,-6,0),C(4,-9,6)\text{ and }D(7,-3,8).$$
$$AB=\sqrt{(2-5)^2+(-6-0)^2+(0-2)^2}$$
$$=\sqrt{9+36+4}$$
$$=\sqrt{49}$$
$$=7$$
$$BC=\sqrt{(4-2)^2+(-9-(-6))^2+(6-0)^2}$$
$$=\sqrt{4+9+36}$$
$$CD=\sqrt{(7-4)^2+(-3-(-9))^2+(8-6)^2}$$
$$DA=\sqrt{(5-7)^2+(0-(-3))^2+(2-8)^2}$$
$$AC=\sqrt{(4-5)^2+(-9-0)^2+(6-2)^2}$$
$$=\sqrt{1+81+16}$$
$$=\sqrt{98}$$
$$BD=\sqrt{(7-2)^2+(-3-(-6))^2+(8-0)^2}$$
$$=\sqrt{25+9+64}$$
$$\textbf{Step -2: Find the correct option.}$$
$$\because AB=BC=CD=DA$$
$$\therefore\text{All sides are equal.}$$
$$\text{and }AC=BD$$
$$\text{They must be its diagonal and they are also equal.}$$
$$\text{So, the figure formed from the given vertices is a square.}$$
$$\textbf{Hence , the correct option is A.}$$
The equation of plane is
$$\dfrac{x}{A}+\dfrac{y}{B}+\dfrac{z}{C}=1$$ ……. (1)
Using centroid formula in XYZ plane is
$$\left( x=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},y=\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3},z=\dfrac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3} \right)$$
Then,
$$\left( \dfrac{A}{3},\dfrac{B}{3},\dfrac{C}{3} \right)=\left( p,q,r \right)$$
Therefore,
$$A=3p,\,\,\,B=3q,\,\,C=3r$$
Put the value of $$A,B$$ and $$C$$ in equation of plane (1) and we get
$$\dfrac{x}{A}+\dfrac{y}{B}+\dfrac{z}{C}=1$$
$$\dfrac{x}{3p}+\dfrac{y}{3q}+\dfrac{z}{3r}=1$$
Hence,
$$\dfrac{x}{p}+\dfrac{y}{q}+\dfrac{z}{r}=3$$
option (D) is correct answer.
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