Step 1: Identify the roots and relationships\n \nLet $\alpha$ and $\beta$ be the roots of the quadratic equation\n\n\[\n\npx^2 + qx - r = 0.\n\n\]\n \nBy Vieta's formulas:\n\n\[\n\n\alpha + \beta = -\frac{q}{p} \tag{1}\n\n\]\n\n\[\n\n\alpha \beta = -\frac{r}{p} \tag{2}\n\n\]\n \n{Step 2: Use the given condition}\n \nGiven:\n\n\[\n\n\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4}.\n\n\]\n \nThis can be rewritten as:\n\n\[\n\n\frac{\alpha + \beta}{\alpha \beta} = \frac{3}{4}.\n\n\]\n \nSubstituting from (1) and (2):\n\n\[\n\n\frac{-\frac{q}{p}}{-\frac{r}{p}} = \frac{3}{4}\n\n\]\n \n\[\n\n\Rightarrow \frac{q}{r} = \frac{3}{4}. \tag{3}\n\n\]\n \n{Step 3: Establish the G.P. relationship}\n \nSince $p, q, r$ are consecutive terms of a non-constant geometric progression, let the common ratio be $k$:\n\n\[\n\n\frac{q}{p} = k, \quad \frac{r}{q} = k.\n\n\]\n \nFrom (3):\n\n\[\n\n\frac{r}{q} = \frac{4}{3}.\n\n\]\n \nHence:\n\n\[\n\nr = \frac{4}{3}q. \tag{4}\n\n\]\n \n{Step 4: Express $p$ in terms of $q$}\n \nFrom $\frac{q}{p} = k$:\n\n\[\n\np = \frac{q}{k}. \tag{5}\n\n\]\n \n{Step 5: Find $(\alpha - \beta)^2$}\n \nWe use the identity:\n\n\[\n\n(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta.\n\n\]\n \nSubstituting from (1) and (2):\n\n\[\n\n(\alpha - \beta)^2 = \left(-\frac{q}{p}\right)^2 - 4\left(-\frac{r}{p}\right)\n\n\]\n \n\[\n\n= \frac{q^2}{p^2} + \frac{4r}{p}.\n\n\]\n \n{Step 6: Substitute the value of $r$}\n \nUsing $r = \frac{4}{3}q$:\n\n\[\n\n(\alpha - \beta)^2 = \frac{q^2}{p^2} + \frac{16q}{3p}.\n\n\]\n \n{Step 7: Express everything in terms of $k$}\n \nSince $p = \frac{q}{k}$:\n\n\[\n\n(\alpha - \beta)^2\n\n= \frac{q^2}{\left(\frac{q}{k}\right)^2}\n\n+ \frac{16q}{3\left(\frac{q}{k}\right)}\n\n\]\n \n\[\n\n= k^2 + \frac{16k}{3}.\n\n\]\n \n{Step 8: Find the value of $k$}\n \nFrom earlier:\n\n\[\n\n\frac{q}{p} = \frac{4}{3} \Rightarrow k = \frac{4}{3}.\n\n\]\n \nSubstituting:\n\n\[\n\n(\alpha - \beta)^2\n\n= \left(\frac{4}{3}\right)^2 + \frac{16}{3}\cdot\frac{4}{3}\n\n= \frac{16}{9} + \frac{64}{9}\n\n= \frac{80}{9}.\n\n\]\n\n