The absolute difference of the coefficients of $$x^{10}$$ and $$x^7$$ in the expansion of $$\left(2x^2 + \frac{1}{2x}\right)^{11}$$ is equal to

Given: Find the absolute difference of the coefficients of $$x^{10}$$ and $$x^7$$ in the expansion of $$\left(2x^2 + \frac{1}{2x}\right)^{11}$$.

General term of the expansion:

$$T_{r+1} = \binom{11}{r}(2x^2)^{11-r}\left(\frac{1}{2x}\right)^r = \binom{11}{r} \cdot 2^{11-r} \cdot x^{2(11-r)} \cdot 2^{-r} \cdot x^{-r}$$

$$= \binom{11}{r} \cdot 2^{11-2r} \cdot x^{22-3r}$$

Coefficient of $$x^{10}$$:

Set $$22 - 3r = 10$$, which gives $$r = 4$$.

$$\text{Coefficient} = \binom{11}{4} \cdot 2^{11-8} = 330 \times 2^3 = 330 \times 8 = 2640$$

Coefficient of $$x^7$$:

Set $$22 - 3r = 7$$, which gives $$r = 5$$.

$$\text{Coefficient} = \binom{11}{5} \cdot 2^{11-10} = 462 \times 2^1 = 462 \times 2 = 924$$

Absolute difference:

$$|2640 - 924| = 1716$$

Now, $$12^3 - 12 = 1728 - 12 = 1716$$ ✓

The correct answer is Option D: $$12^3 - 12$$.