Let the coefficients of third, fourth and fifth terms in the expansion of x + a x 2 n ,   x ≠ 0 ,…

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

Hence, the general term in the expansion of ${\left(x+\frac{a}{x^2}\right)}^n$ is $T_{r+1}={}^{n}C_r(x{)}^{n-r}{\left(\frac{a}{x^2}\right)}^r$

$\Rightarrow T_{r+1}={}^{n}C_r{a}^r{x}^{n-3r}$

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

$\Rightarrow \left({}^{n}C_2{a}^2\right):\left({}^{n}C_3{a}^3\right):{}^{n}C_4{a}^4=12:8:3$

$\Rightarrow \frac{{}^{n}C_2{a}^2}{{}^{n}C_3{a}^3}=\frac{12}{8}$ and $\frac{{}^{n}C_3{a}^3}{{}^{n}C_4{a}^4}=\frac{8}{3}$

Using $\frac{{}^{n}C_r}{{}^{n}C_{r+1}}=\frac{r+1}{n-r},$

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

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

$\Rightarrow 4n-12=3n-6$

$\Rightarrow n=6$ and $a=\frac{1}{2}.$

For term independent of $x\Rightarrow n=3r$

$\Rightarrow r=2.$

$\therefore$ The coefficient is${}^{6}C_2{\left(\frac{1}{2}\right)}^2=\frac{15}{4}.$

Hence, the nearest integer is $4.$