If the constant term in the expansion of $$\left(3x^3 - 2x^2 + \frac{5}{x^5}\right)^{10}$$ is $$2^k \cdot l$$, where $$l$$ is an odd integer, then the value of $$k$$ is equal to

We need the constant term in the expansion of $$\left(3x^3 - 2x^2 + \frac{5}{x^5}\right)^{10}$$.

Using the multinomial theorem, the general term is:

$$\frac{10!}{a!\, b!\, c!}(3x^3)^a(-2x^2)^b\left(\frac{5}{x^5}\right)^c$$

where $$a + b + c = 10$$.

The power of $$x$$ is $$3a + 2b - 5c = 0$$ (for the constant term).

From $$a + b + c = 10$$, we get $$c = 10 - a - b$$. Substituting into the exponent condition:

$$3a + 2b - 5(10 - a - b) = 0$$

$$3a + 2b - 50 + 5a + 5b = 0$$

$$8a + 7b = 50$$

Finding non-negative integer solutions: $$b = \frac{50 - 8a}{7}$$.

For $$a = 1$$: $$b = \frac{42}{7} = 6$$, $$c = 3$$ $$\checkmark$$

No other non-negative integer solutions exist (checking $$a = 0, 2, 3, \ldots$$ yields non-integer values of $$b$$).

The constant term is:

$$\frac{10!}{1!\, 6!\, 3!} \times 3^1 \times (-2)^6 \times 5^3$$

$$= 840 \times 3 \times 64 \times 125$$

$$= 840 \times 24000 = 20160000$$

Factoring out powers of 2: $$20160000 = 2016 \times 10000 = 2016 \times 10^4$$.

$$2016 = 2^5 \times 63$$ and $$10^4 = 2^4 \times 5^4$$.

So $$20160000 = 2^9 \times 63 \times 5^4 = 2^9 \times 63 \times 625 = 2^9 \times 39375$$.

Since $$39375$$ is odd, we have $$k = 9$$.

The answer is Option D: 9.