If the coefficients of x 7 in a x 2 + 1 2 b x 11 and x - 7 in a x - 1 3 b x 2 11 are equal, then

The coefficient of $x^7$ in ${\left(a{x}^2+\frac{1}{2bx}\right)}^{11}$ and ${\left(ax-\frac{1}{3b{x}^2}\right)}^{11}$ are equal, then

$T_{r+1}{=}^{11}{C}_r{\left(a{x}^2\right)}^{11-r}{\left(\frac{1}{2bx}\right)}^r$

${=}^{11}{C}_r{\left(a\right)}^{11-r}{\left(\frac{1}{2b}\right)}^r{x}^{22-3r}$

Now solving $22-3r=7\Rightarrow r=5$

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

$T_{r+1}{=}^{11}{C}_r{\left(ax\right)}^{11-r}{\left(-\frac{1}{3b{x}^2}\right)}^r$

${=}^{11}{C}_r{a}^{11-r}{\left(-\frac{1}{3b}\right)}^r{x}^{11-3r}$

Now solving $11-3r=-7\Rightarrow r=6$

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

$\therefore { }^{11}{C}_5{\left(a\right)}^6{\left(\frac{1}{2b}\right)}^5{=}^{11}{C}_6 a^5{\left(-\frac{1}{3b}\right)}^6$

$\Rightarrow 3^{6}ab=32$

$\Rightarrow 729ab=32$

Hence this is the correct option.