If the coefficients of $$x^7$$ in $$\left(ax^2 + \dfrac{1}{2bx}\right)^{11}$$ and $$x^{-7}$$ in $$\left(ax - \dfrac{1}{3bx^2}\right)^{11}$$ are equal, then

We need to find the relation between $$a$$ and $$b$$ given that the coefficient of $$x^7$$ in $$\left(ax^2 + \dfrac{1}{2bx}\right)^{11}$$ equals the coefficient of $$x^{-7}$$ in $$\left(ax - \dfrac{1}{3bx^2}\right)^{11}$$.

To determine the coefficient of $$x^7$$ in the expansion of $$\left(ax^2 + \dfrac{1}{2bx}\right)^{11}$$, consider the general term $$T_{r+1} = \binom{11}{r}(ax^2)^{11-r}\left(\dfrac{1}{2bx}\right)^r = \binom{11}{r}\dfrac{a^{11-r}}{(2b)^r} \cdot x^{22-3r}\,.$$ Requiring the exponent to be 7 gives $$22 - 3r = 7 \Rightarrow r = 5$$, so the coefficient is $$\binom{11}{5}\dfrac{a^6}{(2b)^5} = \dfrac{462 \cdot a^6}{32b^5}\,.$$

Similarly, for the coefficient of $$x^{-7}$$ in the expansion of $$\left(ax - \dfrac{1}{3bx^2}\right)^{11}$$, the general term is $$T_{r+1} = \binom{11}{r}(ax)^{11-r}\left(\dfrac{-1}{3bx^2}\right)^r = \binom{11}{r}\dfrac{a^{11-r}(-1)^r}{(3b)^r} \cdot x^{11-3r}\,. $$ Setting $$11 - 3r = -7$$ gives $$r = 6$$ and the coefficient becomes $$\binom{11}{6}\dfrac{a^5 \cdot (-1)^6}{(3b)^6} = \dfrac{462 \cdot a^5}{729b^6}\,.$$

Equating these two coefficients yields

$$\frac{462 \cdot a^6}{32b^5} = \frac{462 \cdot a^5}{729b^6}$$

Cancelling $$462 \cdot a^5 / b^5$$ from both sides leads to

$$\frac{a}{32} = \frac{1}{729b}$$

$$729ab = 32$$

Therefore, the required relation is $$729ab = 32$$, which corresponds to Option A.