Let the coefficients of third, fourth and fifth terms in the expansion of $\left(x+\frac{a}{x^{2}}\right)^{n}, x \neq 0$, be in the ratio 12: $8: 3$. Then the term independent of $x$ in the expansion, is equal to
$\mathrm{T}_{\mathrm{r}+1}={ }^{n} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{\mathrm{n}-\mathrm{r}}\left(\frac{\mathrm{a}}{\mathrm{x}^{2}}\right)^{\mathrm{r}}$
$={ }^{n} \mathrm{C}_{\mathrm{r}} \mathrm{a}^{\mathrm{r}} \mathrm{x}^{\mathrm{n}-3 \mathrm{r}}$
${ }^{\mathrm{n}} \mathrm{C}_{2} \mathrm{a}^{2}:{ }^{n} \mathrm{C}_{3} \mathrm{a}^{3}:{ }^{n} \mathrm{C}_{4} \mathrm{a}^{4}=12: 8: 3$
After solving
$\mathrm{n}=6, \mathrm{a}=\frac{1}{2}$
For term independent of ' $x$ ' $\Rightarrow n=3 r$
$\mathrm{r}=2$
$\therefore$ Coefficient is ${ }^{6} \mathrm{C}_{2}\left(\frac{1}{2}\right)^{2}=\frac{15}{4}$
Nearest integer is 4