If the function $$f(x) = \begin{cases} (1+|\cos x|)^{\frac{\lambda}{|\cos x|}}, & 0 < x < \frac{\pi}{2} \\ \mu, & x = \frac{\pi}{2} \\ e^{\frac{\cot 6x}{\cot 4x}}, & \frac{\pi}{2} < x < \pi \end{cases}$$ is continuous at $$x = \frac{\pi}{2}$$, then $$9\lambda + 6\log_e \mu + \mu^6 - e^{6\lambda}$$ is equal to

For continuity at $$x = \frac{\pi}{2}$$, we need the left-hand limit, right-hand limit, and function value to be equal.

Left-hand limit ($$x \to \frac{\pi}{2}^-$$):

As $$x \to \frac{\pi}{2}^-$$, $$|\cos x| \to 0^+$$. Let $$u = |\cos x|$$:

$$(1 + u)^{\lambda/u} \to e^{\lambda} \quad \text{(standard limit)}$$

Right-hand limit ($$x \to \frac{\pi}{2}^+$$):

Let $$x = \frac{\pi}{2} + h$$ where $$h \to 0^+$$:

$$\cot 6x = \cot(3\pi + 6h) = \cot(6h) \quad (\text{period } \pi)$$

$$\cot 4x = \cot(2\pi + 4h) = \cot(4h) \quad (\text{period } \pi)$$

$$\frac{\cot 6h}{\cot 4h} = \frac{\cos 6h \cdot \sin 4h}{\sin 6h \cdot \cos 4h} \to \frac{4h}{6h} = \frac{2}{3} \quad \text{as } h \to 0$$

So the right-hand limit is $$e^{2/3}$$.

Continuity condition:

$$e^{\lambda} = \mu = e^{2/3}$$

$$\lambda = \frac{2}{3}, \quad \mu = e^{2/3}$$

Computing the expression:

$$9\lambda + 6\log_e \mu + \mu^6 - e^{6\lambda}$$

$$= 9 \cdot \frac{2}{3} + 6 \cdot \frac{2}{3} + (e^{2/3})^6 - e^{6 \cdot 2/3}$$

$$= 6 + 4 + e^4 - e^4 = 10$$

Therefore, the correct answer is Option D: $$\mathbf{10}$$.