Use Vieta’s formulas for the quadratic

$px^2+qx-r=0$

with roots $\alpha,\beta$:

$\alpha+\beta=-\frac{q}{p}, \qquad \alpha\beta=-\frac{r}{p}$

Use the given condition

$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{3}{4}$

Substituting Vieta’s values,

$\frac{-q/p}{-r/p}=\frac{q}{r}=\frac{3}{4}$

Hence,

$q:r=3:4$

Use the GP condition

Since $p,q,r$ are consecutive terms of a non-constant G.P., we have

$q^2=pr$

Also from $q:r=3:4$, let

$q=3k,\qquad r=4k$

Then

$q^2=pr \implies 9k^2=p(4k) \implies p=\frac{9k}{4}$

So

$p:q:r=\frac{9}{4}:3:4$

Multiplying by $4$,

$p:q:r=9:12:16$

Form the quadratic (up to a nonzero constant factor)

$9x^2+12x-16=0$

Its roots are still $\alpha,\beta$.

Compute $(\alpha-\beta)^2$

Using

$(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$

For the equation $9x^2+12x-16=0$,

$\alpha+\beta=-\frac{12}{9}=-\frac{4}{3}, \qquad \alpha\beta=-\frac{-16}{9}= -\frac{16}{9}?$

Careful: for $ax^2+bx+c=0$, product of roots is $c/a$. Here $c=-16$, so

$\alpha\beta=\frac{-16}{9}=-\frac{16}{9}$

Therefore,

$(\alpha-\beta)^2=\left(-\frac{4}{3}\right)^2-4\left(-\frac{16}{9}\right)$

$=\frac{16}{9}+\frac{64}{9}=\frac{80}{9}$

Check options

$\frac{80}{9}$

matches Option D.