If the coefficients of x ^ 7 in (a x ^ 2 + (1)/(2bx)) ^ 11 and x ^...

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

${T}_{r + 1}{=}^{11}{C}_{r}{\left(a{x}^{2}\right)}^{11 - r}{\left(\dfrac{1}{2bx}\right)}^{r}$

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

Now solving $22 - 3r = 7 ⇒ r = 5$

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

${T}_{r + 1}{=}^{11}{C}_{r}{\left(ax\right)}^{11 - r}{\left(- \dfrac{1}{3b{x}^{2}}\right)}^{r}$

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

Now solving $11 - 3r = - 7 ⇒ r = 6$

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

$∴ {}^{11}{C}_{5}{\left(a\right)}^{6}{\left(\dfrac{1}{2b}\right)}^{5}{=}^{11}{C}_{6}{a}^{5}{\left(- \dfrac{1}{3b}\right)}^{6}$

$⇒ {3}^{6}ab = 32$

$⇒ 729ab = 32$

Hence this is the correct option.