Let $\mathrm{\alpha }$ be the constant term in the binomial expansion of ${\left(\sqrt{\mathrm{x}}-\frac{{\displaystyle 6}}{{\displaystyle {\mathrm{x}}^{\frac{{\displaystyle 3}}{{\displaystyle 2}}}}}\right)}^n,\mathrm{n}\le 15$. If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of ${\mathrm{x}}^{-\mathrm{n}}$ is $\lambda \alpha ,$then $\mathrm{\lambda }$ is equal to _____________.

$$\begin{gathered} \begin{array}{l}{T}_{r-1}{=}^n{C}_r{\left(x^{\frac{1}{2}}\right)}^{n-r}{\left(-6x^{\frac{-3}{2}}\right)}^r\\ =(-6{)}^r{\cdot }^n{C}_r\cdot x^{\frac{n-r}{2}}\cdot x^{\frac{-3r}{2}}\end{array} \\ n=4r\text{ for constant term so }n\text{ is divisible by }4\text{. } \\ \begin{array}{l}{ }^n{C}_{\frac{n}{4}}(-6{)}^{\frac{n}{4}}=\alpha \\ sum of the coefficients ={\left(-5\right)}^n \left(\sin ce by putting x=1\right)\\ now (-5{)}^n{-}^n{C}_{\frac{n}{4}}(-6{)}^{n/4}=649\\ n=4\\ \alpha =-24\end{array} \end{gathered}$$
$$\begin{gathered} \text{ For coefficient of }{x}^{-4} \\ \begin{array}{l}\frac{n-4r}{2}=-4\\ n=4r-8\Rightarrow r=3\\ \lambda (-24)=(-6{)}^3{\cdot }^4{C}_3\\ \lambda =36\end{array} \end{gathered}$$