If the function $$f: (-\infty, -1] \rightarrow [a, b]$$ defined by $$f(x) = e^{x^3 - 3x + 1}$$ is one-one and onto, then the distance of the point $$P(2b + 4, a + 2)$$ from the line $$x + e^{-3}y = 4$$ is:

Let $$g(x) = x^3 - 3x + 1$$, so $$f(x) = e^{g(x)}$$. 

$$g'(x) = 3x^2 - 3 = 3(x-1)(x+1).$$ 

On $$(-\infty, -1)$$ we have $$g'(x) > 0$$ , so $$g$$ is strictly increasing. 

At $$x = -1$$, $$g(-1) = -1 + 3 + 1 = 3$$, giving a local maximum of $$g$$. As $$x \to -\infty$$, $$g(x) \to -\infty$$. 

Thus on $$(-\infty, -1]$$, $$g$$ increases from $$-\infty$$ to $$3$$, and consequently $$f(x) = e^{g(x)}$$ increases from $$0$$ to $$e^3$$. 

Therefore the range of $$f$$ on this domain is $$(0, e^3]$$, which implies $$a = 0$$ and $$b = e^3$$.

 $$P(2b + 4, a + 2) = P(2e^3 + 4, 2).$$

 $$x + e^{-3}y - 4 = 0$$. 

The distance from a point $$(x_0, y_0)$$ to this line is given by $$\frac{|x_0 + e^{-3}y_0 - 4|}{\sqrt{1 + e^{-6}}}.$$

Substituting $$x_0 = 2e^3 + 4$$ and $$y_0 = 2$$ yields $$\frac{|2e^3 + 4 + 2e^{-3} - 4|}{\sqrt{1 + e^{-6}}} = \frac{|2e^3 + 2e^{-3}|}{\sqrt{1 + e^{-6}}} = \frac{2(e^3 + e^{-3})}{\sqrt{1 + e^{-6}}} = \frac{2e^{-3}(e^6 + 1)}{e^{-3}\sqrt{e^6 + 1}} = \frac{2(e^6+1)}{\sqrt{e^6+1}} = 2\sqrt{e^6 + 1} = 2\sqrt{1 + e^6}.$$