The number of positive integers k such that the constant term in the binomial expansion of ${\left(2{\mathrm{x}}^3+\frac{3}{{\mathrm{x}}^{\mathrm{k}}}\right)}^{12},\mathrm{x}\ne 0\text{ is }{2}^8\cdot \mathrm{\ell }$, where l is an odd integer, is __________.

$\mathrm{In} {\left(\frac{3}{{\mathrm{x}}^{\mathrm{k}}}+2{\mathrm{x}}^3\right)}^{12}$
$\begin{array}{l}{\mathrm{t}}_{\mathrm{r}+1}{=}^{12}{\mathrm{C}}_{\mathrm{r}}{\left(2{\mathrm{x}}^3\right)}^{\mathrm{r}}{\left(\frac{3}{{\mathrm{x}}^{\mathrm{k}}}\right)}^{12-\mathrm{r}}\\ {\mathrm{x}}^{3\mathrm{r}-(12-\mathrm{r})\mathrm{k}}\to \text{ constant }\\ \begin{array}{r}\therefore 3\mathrm{r}-12\mathrm{k}+\mathrm{rk}=0\\ \Rightarrow \mathrm{k}=\frac{3\mathrm{r}}{12-\mathrm{r}}\end{array}\\ \therefore \text{ possible values of }\mathrm{r}\text{ are }3,6,8,9,10\\ \text{ and corresponding values of }\mathrm{k}\text{ are }1,3,6,9,15\\ { }^{12}{\mathrm{C}}_{\mathrm{r}}=220,924,495,220,66\\ \therefore \text{ possible values of }\mathrm{k}\text{ for which we will get }{2}^8\text{ are 3,6}\end{array}$