Let α be the constant term in the binomial expansion of ( x -...

Given,

${\left(\sqrt{x} - \dfrac{6}{x\dfrac{3}{2}}\right)}^{n}, n \le 15$

Now ${r}^{th}$ term is given by

${T}_{r + 1} = Cr n {\left(x\dfrac{1}{2}\right)}^{n - r}{\left(- 6x\dfrac{- 3}{2}\right)}^{r}$

$⇒ {T}_{r + 1} = {\left(- 6\right)}^{r} \cdot Cr n \cdot x\dfrac{n - r}{2} \cdot x\dfrac{- 3r}{2}$

$⇒ {T}_{r + 1} = {\left(- 6\right)}^{r} \cdot Cr n \cdot x\dfrac{n - 4r}{2}$

Now we have to take, $n = 4r$ for constant term so $n$ is divisible by $4 .$

So, $\alpha = C\dfrac{n}{4} n \left(- 6\right)\dfrac{n}{4}$

And sum of all coefficients will be ${\left(\sqrt{1} - \dfrac{6}{1\dfrac{3}{2}}\right)}^{n} = {\left(- 5\right)}^{n}$

Also given, ${\left(- 5\right)}^{n} - C\dfrac{n}{4} n \left(- 6\right)\dfrac{n}{4} = 649$

Now by hit and trial we get, $n = 4$

Hence, $\alpha = C1 4 \left(- 6\right)\dfrac{4}{4} = - 24$

Now for coefficient of ${x}^{- 4}$

$\dfrac{n - 4r}{2} = - 4$

$⇒ n = 4r - 8 ⇒ r = 3$

So, $\lambda\left(\alpha\right) = {\left(- 6\right)}^{3} \cdot C3 4$

$⇒ \lambda\left(- 24\right) = {\left(- 6\right)}^{3} \cdot C3 4$

$⇒ \lambda = 36$