The term independent of $$x$$ in the expression of $$(1 - x^2 + 3x^3)\left(\frac{5}{2}x^3 - \frac{1}{5x^2}\right)^{11}$$, $$x \neq 0$$ is

We need the term independent of $$x$$ in:

$$\left(1 - x^2 + 3x^3\right)\left(\frac{5}{2}x^3 - \frac{1}{5x^2}\right)^{11}$$

Find the general term of the binomial expansion:

The general term of $$\left(\frac{5}{2}x^3 - \frac{1}{5x^2}\right)^{11}$$ is:

$$T_{r+1} = \binom{11}{r}\left(\frac{5}{2}\right)^{11-r}\left(-\frac{1}{5}\right)^r x^{3(11-r) - 2r} = \binom{11}{r}\left(\frac{5}{2}\right)^{11-r}\left(-\frac{1}{5}\right)^r x^{33-5r}$$

Identify which terms contribute to the constant term in the product:

When multiplied by $$(1 - x^2 + 3x^3)$$, we need:

• From $$1 \cdot T_{r+1}$$: need $$33 - 5r = 0 \Rightarrow r = 33/5$$ (not an integer, no contribution)
• From $$(-x^2) \cdot T_{r+1}$$: need $$33 - 5r + 2 = 0 \Rightarrow 5r = 35 \Rightarrow r = 7$$ $$\checkmark$$
• From $$(3x^3) \cdot T_{r+1}$$: need $$33 - 5r + 3 = 0 \Rightarrow 5r = 36 \Rightarrow r = 36/5$$ (not an integer, no contribution)

Compute the coefficient for $$r = 7$$:

$$T_8 = \binom{11}{7}\left(\frac{5}{2}\right)^{4}\left(-\frac{1}{5}\right)^{7}$$

$$= 330 \times \frac{625}{16} \times \frac{(-1)}{78125}$$

$$= 330 \times \frac{-625}{16 \times 78125}$$

$$= 330 \times \frac{-1}{16 \times 125}$$

$$= \frac{-330}{2000} = \frac{-33}{200}$$

Multiply by the coefficient $$(-1)$$ from the $$(-x^2)$$ term:

$$\text{Constant term} = (-1) \times \frac{-33}{200} = \frac{33}{200}$$

The correct answer is Option B: $$\dfrac{33}{200}$$.