Let α and β be the roots of the equation p x ^ 2 + qx - r = 0 ,...

Given: $\alpha\text{and}\beta$ be the roots of equation $p{x}^{2} + qx - r = 0$

So, $\alpha + \beta = - \dfrac{q}{p}, \alpha\beta = - \dfrac{r}{p}$

It is given that, $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{3}{4}$

$⇒ \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{3}{4}$

$⇒ \dfrac{q}{r} = \dfrac{3}{4}$

Now, $p, q, r$ are in GP

So, common ratio of this GP will be $\dfrac{q}{p} = \dfrac{r}{q} = \dfrac{4}{3}$

$⇒ p{x}^{2} + qx - r = 0$

$⇒ {x}^{2} + \dfrac{q}{p}x - \dfrac{r}{p} = 0$

$⇒ {x}^{2} + \dfrac{4}{3}x - {\left(\dfrac{4}{3}\right)}^{2} = 0$

$⇒ 9{x}^{2} + 12x - 16 = 0$

$⇒ {\left(\alpha - \beta\right)}^{2} = {\left(\alpha + \beta\right)}^{2} - 4\alpha\beta$

$⇒ {\left(\alpha - \beta\right)}^{2} = {\left(\dfrac{- 12}{9}\right)}^{2} - 4\left(\dfrac{- 16}{9}\right)$

$⇒ {\left(\alpha - \beta\right)}^{2} = \dfrac{16}{9} + \dfrac{64}{9}$

$⇒ {\left(\alpha - \beta\right)}^{2} = \dfrac{80}{9}$