1. General term of the expansion

For $\left(\sqrt{x}-\frac{6}{x^{3/2}}\right)^n=\left(x^{1/2}-6x^{-3/2}\right)^n,$ the general term is $T_{r+1}=\binom{n}{r}(x^{1/2})^{n-r}(-6x^{-3/2})^r.$

So, T_{r+1}=\binom{n}{r}(-6)^r x^{\frac{n-r}{2}-\frac{3r}{2}}=inom{n}{r}(-6)^r x^{\frac{n-4r}{2}}.

2. Find the constant term

For the constant term, exponent of $x$ must be zero: $\frac{n-4r}{2}=0 \implies n=4r.$ Thus $n$ must be divisible by $4$.

Given $n\le 15$, possible values are $n=4,8,12.$

Let the constant term be $\alpha=\binom{n}{n/4}(-6)^{n/4}.$

3. Use the sum of coefficients condition

The sum of all coefficients in the expansion is obtained by putting $x=1$: $\left(1-6\right)^n=(-5)^n.$

The sum of coefficients of the remaining terms means: $\text{(sum of all coefficients)}-\text{(constant term coefficient)}=649.$ So, $(-5)^n-\alpha=649.$

Now test possible values of $n$.

• For $n=4$: $\alpha=\binom41(-6)=-24,$ $(-5)^4-\alpha=625-(-24)=649.$ Works.

• For $n=8$: $\alpha=\binom82(-6)^2=28\cdot 36=1008,$ $(-5)^8-\alpha=390625-1008\neq 649.$

• For $n=12$: clearly far too large, so it does not work.

Hence, $n=4, \qquad \alpha=-24.$

4. Find the coefficient of $x^{-n}$

We need exponent of $x$ to be $-n$: $\frac{n-4r}{2}=-n.$ Solve: $n-4r=-2n \implies 4r=3n \implies r=\frac{3n}{4}.$

For $n=4$, $r=3.$ So the coefficient of $x^{-n}=x^{-4}$ is $\binom43(-6)^3=4(-216)=-864.$

Given this coefficient is $\lambda\alpha$, $-864=\lambda(-24).$ Therefore, $\lambda=\frac{-864}{-24}=36.$

5. Final answer

$\boxed{36}$