Let $$[t]$$ denote the greatest integer $$\le t$$. If the constant term in the expansion of $$\left(3x^2 - \dfrac{1}{2x^5}\right)^7$$ is $$\alpha$$ then $$[\alpha]$$ is equal to ______.

To find $$[\alpha]$$, we need to determine the constant term (the term where the power of $$x$$ is $$0$$) in the expansion of $$(3x^2 - \frac{1}{2x^5})^7$$.

The general term $$T_{r+1}$$ in the expansion of $$(a + b)^n$$ is $$^nC_r \cdot a^{n-r} \cdot b^r$$.

For $$(3x^2 - \frac{1}{2x^5})^7$$:

$$T_{r+1} = ^7C_r \cdot (3x^2)^{7-r} \cdot \left(-\frac{1}{2x^5}\right)^r$$

Combine the $$x$$ terms to find the total exponent:

$$T_{r+1} = ^7C_r \cdot 3^{7-r} \cdot \left(-\frac{1}{2}\right)^r \cdot \frac{x^{2(7-r)}}{x^{5r}}$$

$$x \text{ exponent} = 14 - 2r - 5r = 14 - 7r$$

Set the exponent to $$0$$:

$$14 - 7r = 0 \implies \mathbf{r = 2}$$

Plug $$r = 2$$ back into the coefficient part:

$$\alpha = ^7C_2 \cdot 3^{7-2} \cdot \left(-\frac{1}{2}\right)^2$$

$$\alpha = 21 \cdot 3^5 \cdot \frac{1}{4}$$

$$\alpha = 21 \cdot 243 \cdot 0.25 = \frac{5103}{4} = \mathbf{1275.75}$$

The problem asks for $$[\alpha]$$, which is the greatest integer $$\le \alpha$$:

$$[1275.75] = \mathbf{1275}$$