Lets $\alpha,\beta$ and $\gamma$ are the roots of $ax^{3}+bx^{2}+cx+d=0,$ then
$\alpha+\beta+\gamma = \dfrac{-b}{a}$
$\alpha\beta+\beta\gamma+\alpha\gamma = \dfrac{c}{a}$
$\alpha\beta\gamma = \dfrac{-d}{a}$
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Lets say roots of the equation $x^{3}-px^{2}+qx-r = 0 $ are $\alpha,\beta$ and $\gamma$ are in $H.P.$
So, $\dfrac{1}{\alpha},\dfrac{1}{\beta},\dfrac{1}{\gamma}$ are in $A.P.$
We know that if $A,B,C$ are in $A.P.$ then $2B=A+C$
$\implies\dfrac{2}{\beta}=\dfrac{1}{\alpha}+\dfrac{1}{\gamma}$
$\implies\dfrac{2}{\beta}=\dfrac{\gamma+\alpha}{\alpha\gamma}$
$\implies 2\alpha\gamma = \beta\gamma+\alpha \beta\rightarrow (1)$
Now,
$\alpha+\beta+\gamma = \dfrac{-b}{a} = \dfrac{-(-p)}{1}=p\rightarrow(2)$
$\alpha\beta+\beta\gamma+\alpha\gamma = \dfrac{c}{a}=\dfrac{q}{1}=q\rightarrow(3)$
$\alpha\beta\gamma = \dfrac{-d}{a}=\dfrac{-(-r)}{1}=r\rightarrow(4)$
From the equation $(1)$ and $(3),$ we get
$\underbrace{\alpha\beta+\beta\gamma}+\alpha\gamma =q$
$\implies 2\alpha\gamma + \alpha\gamma=q$
$\implies 3\alpha\gamma =q$
$\implies \alpha\gamma =\dfrac{q}{3}\rightarrow(5)$
From the equation $(4)$ and $(5),$ we get
$\alpha\beta\gamma = r $
$\implies\underbrace{\alpha\gamma}\beta = r $
$\implies\dfrac{q}{3}\beta = r $
$\implies\beta=\dfrac{3r}{q}$
From the equation $(2),$ we get
$\alpha+\beta +\gamma = p$
$\implies \alpha +\gamma = \dfrac{pq-3r}{q}\rightarrow(6)$
From the equation $(1),$ we get
$2\alpha\gamma = \beta\gamma+\alpha \beta$
$\implies \underbrace{2\alpha\gamma} =\underbrace{ \beta}\left[\gamma+\alpha\right]$
$\implies \dfrac{2q}{3}=\dfrac{3r}{q}\left[ \alpha+\gamma\right]$
$\implies \alpha + \gamma=\dfrac{2q^{2}}{9r}\rightarrow (7)$
Now, from the equation $(6)$ and $(7),$ we get
$\alpha +\gamma = \dfrac{pq-3r}{q}=\dfrac{2q^{2}}{9r}$
$\implies \dfrac{pq-3r}{q}=\dfrac{2q^{2}}{9r}$
$\implies 9pqr-27r^{2}=2q^{3}$
$\implies 27r^{2}-9pqr+2q^{3}=0$
So, the correct answer is $(D)$.