If the coefficients of x^7 in (ax^2+ 1 2bx)^11 and x^-7 in (ax- 1 3bx^2)^11 are equal, then

The coefficient of ${x}^{7}$x7 in ${(a{x}^{2}+\frac{1}{2bx})}^{11}$(ax2+2bx1​)11 and ${(ax-\frac{1}{3b{x}^{2}})}^{11}$(ax−3bx21​)11 are equal, then

$${T}_{r+1}{=}^{11}{C}_{r}{(a{x}^{2})}^{11-r}{(\frac{1}{2bx})}^{r}$$Tr+1​=11Cr​(ax2)11−r(2bx1​)r

$${=}^{11}{C}_{r}{(a)}^{11-r}{(\frac{1}{2b})}^{r}{x}^{22-3r}$$=11Cr​(a)11−r(2b1​)rx22−3r

Now solving $22-3r=7\Rightarrow r=5$22−3r=7⇒r=5

Coefficient of ${x}^{-7}$x−7 in ${(ax-\frac{1}{3b{x}^{2}})}^{11}$(ax−3bx21​)11

$${T}_{r+1}{=}^{11}{C}_{r}{(ax)}^{11-r}{(-\frac{1}{3b{x}^{2}})}^{r}$$Tr+1​=11Cr​(ax)11−r(−3bx21​)r

$${=}^{11}{C}_{r}{a}^{11-r}{(-\frac{1}{3b})}^{r}{x}^{11-3r}$$=11Cr​a11−r(−3b1​)rx11−3r

Now solving $11-3r=-7\Rightarrow r=6$11−3r=−7⇒r=6

Now equating coefficient of ${x}^{7}$x7 and ${x}^{-7}$x−7 we get,

$$\therefore {}^{11}{C}_{5}{(a)}^{6}{(\frac{1}{2b})}^{5}{=}^{11}{C}_{6}{a}^{5}{(-\frac{1}{3b})}^{6}$$∴11C5​(a)6(2b1​)5=11C6​a5(−3b1​)6

$$\Rightarrow {3}^{6}ab=32$$⇒36ab=32

$$\Rightarrow 729ab=32$$⇒729ab=32

Hence this is the correct option.