If for $$p \neq q \neq 0$$, the function $$f(x) = \frac{7\sqrt[p]{729+x} - 3}{\sqrt[3]{729+qx} - 9}$$ is continuous at $$x = 0$$, then

Given,

$$f(x)=\frac{\sqrt[7]{p(729+x)}-3}{\sqrt[3]{729+qx}-9}$$

For continuity at

$$x=0,$$

the limit must exist.

Substituting $$x=0$$ in denominator,

$$\sqrt[3]{729}-9=9-9=0$$

Hence numerator must also become zero.

Therefore,

$$\sqrt[7]{p(729)}-3=0$$

Since

$$729=3^6,$$

$$\sqrt[7]{3^6p}=3$$

$$3^6p=3^7$$

$$p=3$$

Now evaluate

$$f(0)=\lim_{x\to0}\frac{\sqrt[7]{3(729+x)}-3}{\sqrt[3]{729+qx}-9}$$

Using L’Hôpital’s Rule,

$$f(0)=\lim_{x\to0}\frac{\frac{1}{7}[3(729+x)]^{-6/7}\cdot3}{\frac13(729+qx)^{-2/3}\cdot q}$$

Substituting $$x=0,$$

$$f(0)=\frac{\frac{3}{7(3^7)^{6/7}}}{\frac{q}{3(3^6)^{2/3}}}$$

$$=\frac{\frac{3}{7\cdot3^6}}{\frac{q}{3\cdot3^4}}$$

$$=\frac{1}{7q}$$

Hence,

$$7qf(0)=1$$

Since

$$p=3,$$

$$p^2=9$$

Multiplying by $$9,$$

$$63qf(0)=9=p^2$$

Therefore,

$$63qf(0)-p^2=0$$

Hence, the correct relation is

$$\boxed{63qf(0)-p^2=0}$$.