Let the coefficients of third, fourth and fifth terms in the...

The general term in the expansion of ${\left(a + b\right)}^{n}$ is ${T}_{r + 1} = Cr n {a}^{n - r}{b}^{r} .$

Hence, the general term in the expansion of ${\left(x + \dfrac{a}{{x}^{2}}\right)}^{n}$ is ${T}_{r + 1} = Cr n ( x{)}^{n - r}{\left(\dfrac{a}{{x}^{2}}\right)}^{r}$

$⇒ {T}_{r + 1} = Cr n {a}^{r}{x}^{n - 3r}$

Given the coefficients of third, fourth and fifth terms are in the ratio $12: 8: 3$

$⇒ \left(C2 n {a}^{2}\right): \left(C3 n {a}^{3}\right): C4 n {a}^{4} = 12: 8: 3$

$⇒ \dfrac{C2 n {a}^{2}}{C3 n {a}^{3}} = \dfrac{12}{8}$ and $\dfrac{C3 n {a}^{3}}{C4 n {a}^{4}} = \dfrac{8}{3}$

Using $\dfrac{Cr n}{Cr + 1 n} = \dfrac{r + 1}{n - r},$

$⇒ \dfrac{3}{\left(n - 2\right)a} = \dfrac{3}{2}$ and $\dfrac{4}{\left(n - 3\right)a} = \dfrac{8}{3}$

On dividing the two equations, we get $\dfrac{3\left(n - 3\right)}{4\left(n - 2\right)} = \dfrac{9}{16}$

$⇒ 4n - 12 = 3n - 6$

$⇒ n = 6$ and $a = \dfrac{1}{2} .$

For term independent of $x ⇒ n = 3r$

$⇒ r = 2 .$

$∴$ The coefficient is$C2 6 {\left(\dfrac{1}{2}\right)}^{2} = \dfrac{15}{4} .$

Hence, the nearest integer is $4 .$