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

General term of $\left(2 x^2+\frac{1}{2 x}\right)^{11}$ is : <br/><br/>$$ \begin{aligned} & \mathrm{T}_{\mathrm{r}+1}={ }^{11} \mathrm{C}_r\left(2 x^2\right){ }^{11-r}\left(\frac{1}{2 x}\right)^r \\\\ & ={ }^{11} \mathrm{C}_r 2^{11-r} x^{22-2 r} 2^{-r} x^{-r} \\\\ & ={ }^{11} \mathrm{C}_r 2^{11-r} x^{22-3 r} \end{aligned} $$ <br/><br/>Now, $22-2 r=10$ and $22-3 r=7$ <br/><br/>$$ \begin{array}{ll} \Rightarrow 3 r=12 &&& \Rightarrow 3 r=15 \\\\ \Rightarrow r=4 &&& \Rightarrow r=5 \end{array} $$ <br/><br/>$\therefore$ Coeff. of $x^{10}={ }^{11} \mathrm{C}_4 \cdot 2^{11-8}={ }^{11} \mathrm{C}_4 \times 8$ <br/><br/>Coeff. of $x^7={ }^{11} C_5 \cdot 2^{11-10}={ }^{11} C_4 \times 2$ <br/><br/>Now, required difference <br/><br/>$$ \begin{aligned} & ={ }^{11} \mathrm{C}_4 \times 8-{ }^{11} \mathrm{C}_5 \times 2 \\\\ & =\frac{11 \times 10 \times 9 \times 8 \times 7 !}{4 ! \times 7 !} \times 8-\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 ! \times 2}{5 ! 6 !} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & =\frac{11 \times 10 \times 9 \times 8 \times 8}{24}-\frac{11 \times 10 \times 9 \times 8 \times 7 \times 2}{120} \\\\ & =11 \times 10 \times 8 \times 3-11 \times 3 \times 4 \times 7 \\\\ & =11 \times 3 \times 4[20-7] \\\\ & =11 \times 12 \times 13=(12-1) \times 12 \times(12+1) \\\\ & =12\left(12^2-1\right)=12^3-12 \end{aligned} $$