Step-by-step Solution:

1. Identify the relations from the first quadratic equation. The first equation is $x^2 - px + r = 0$. Let its roots be $\alpha$ and $\beta$. Using Vieta's formulas, we have:

   • Sum of roots: $\alpha + \beta = -(\frac{-p}{1}) = p$ ... (i)
   • Product of roots: $\alpha \beta = \frac{r}{1} = r$ ... (ii)

2. Identify the relations from the second quadratic equation. The second equation is $x^2 - qx + r = 0$. Let its roots be $\frac{\alpha}{2}$ and $2\beta$. Using Vieta's formulas, we have:

   • Sum of roots: $\frac{\alpha}{2} + 2\beta = -(\frac{-q}{1}) = q$ ... (iii)
   • Product of roots: $(\frac{\alpha}{2})(2\beta) = \frac{r}{1} = r$. This simplifies to $\alpha \beta = r$, which is consistent with equation (ii).

3. Solve the system of linear equations for $\alpha$ and $\beta$. We have a system of two linear equations in $\alpha$ and $\beta$ from equations (i) and (iii):

   • $\alpha + \beta = p$ ... (i)
   • $\frac{\alpha}{2} + 2\beta = q$. Multiplying by 2, we get $\alpha + 4\beta = 2q$ ... (iv)

   To solve for $\beta$, subtract equation (i) from equation (iv): $(\alpha + 4\beta) - (\alpha + \beta) = 2q - p$ $3\beta = 2q - p$ $\beta = \frac{2q - p}{3}$

   Now, substitute the value of $\beta$ back into equation (i) to find $\alpha$: $\alpha = p - \beta$ $\alpha = p - \frac{2q - p}{3}$ $\alpha = \frac{3p - (2q - p)}{3}$ $\alpha = \frac{3p - 2q + p}{3}$ $\alpha = \frac{4p - 2q}{3} = \frac{2(2p - q)}{3}$

4. Calculate the value of r. From equation (ii), we know that $r = \alpha \beta$. Substitute the expressions we found for $\alpha$ and $\beta$: $r = \left( \frac{2(2p - q)}{3} \right) \left( \frac{2q - p}{3} \right)$ $r = \frac{2}{9}(2p - q)(2q - p)$

5. Compare the result with the given options. The calculated value of $r$ is $\frac{2}{9}(2p-q)(2q-p)$. This matches option D.

Conclusion:

The correct value of r is $\frac{2}{9}(2p-q)(2q-p)$.