The equation of plane is

${\displaystyle \frac{x}{A}}+{\displaystyle \frac{y}{B}}+{\displaystyle \frac{z}{C}}=1$           …….   (1)

Using centroid formula in XYZ plane is

$(x={\displaystyle \frac{x_1+x_2+x_3}{3}},y={\displaystyle \frac{y_1+y_2+y_3}{3}},z={\displaystyle \frac{z_1+z_2+z_3}{3}})$

Then,

$({\displaystyle \frac{A}{3}},{\displaystyle \frac{B}{3}},{\displaystyle \frac{C}{3}})=(p,q,r)$

Therefore,

$A=3p,B=3q,C=3r$

Put the value of $A,B$ and $C$ in equation of plane (1) and we get

${\displaystyle \frac{x}{A}}+{\displaystyle \frac{y}{B}}+{\displaystyle \frac{z}{C}}=1$

${\displaystyle \frac{x}{3p}}+{\displaystyle \frac{y}{3q}}+{\displaystyle \frac{z}{3r}}=1$

Hence,

${\displaystyle \frac{x}{p}}+{\displaystyle \frac{y}{q}}+{\displaystyle \frac{z}{r}}=3$

option (D) is correct answer.