If $\alpha ,\beta$ are the roots of equation $x^2+px-r=0$ and $\frac{\alpha }{3},3\beta$ are the roots of the equation $x^2+qx-r=0$ then r equals

Given,  

$\therefore \alpha +\beta =-p$ and $\alpha \beta =-r \dots \left(i\right)$

Also,   …(ii)

$\therefore \frac{\alpha }{3}+3\beta =-q\to \alpha +9\beta =-3q$      …(iii)

and $\left(\frac{\alpha }{3}\right)\left(3\beta \right)=-r\Rightarrow \alpha \beta =-r$

So on solving (i) and (iii) we get $\alpha =\frac{3(q-3p)}{8},\beta =\frac{(p-3q)}{8}$

as $\alpha \beta =-r\Rightarrow r=\frac{3(3p-q)(p-3q)}{64}$