If the absolute maximum value of the function $$f(x) = (x^2 - 2x + 7)e^{(4x^3 - 12x^2 - 180x + 31)}$$ in the interval $$[-3, 0]$$ is $$f(\alpha)$$, then

Given,

$$f(x)=(x^2-2x+7)e^{(4x^3-12x^2-180x+31)}$$

on the interval

$$[-3,0]$$

Let

$$g(x)=x^2-2x+7$$

and

$$h(x)=4x^3-12x^2-180x+31$$

Then,

$$f(x)=g(x)e^{h(x)}$$

Differentiate:

$$f'(x)=e^{h(x)}\left[g'(x)+g(x)h'(x)\right]$$

Now,

$$g'(x)=2x-2$$

and

$$h'(x)=12x^2-24x-180$$

$$=12(x-5)(x+3)$$

Therefore,

$$f'(x)=e^{h(x)}\left[(2x-2)+(x^2-2x+7)(12x^2-24x-180)\right]$$

Since

$$e^{h(x)}>0,$$

critical points come from

$$g'(x)+g(x)h'(x)=0$$

Now check important points in

$$[-3,0]$$

At

$$x=-3,$$

$$h(-3)=4(-27)-12(9)-180(-3)+31$$

$$=-108-108+540+31$$

$$=355$$

and

$$g(-3)=9+6+7=22$$

Hence,

$$f(-3)=22e^{355}$$

At

$$x=0,$$

$$h(0)=31$$

and

$$g(0)=7$$

Thus,

$$f(0)=7e^{31}$$

Clearly,

$$22e^{355}\gg7e^{31}$$

Also,

$$h'(x)=12(x-5)(x+3)<0$$

for

$$-3<x<0$$

Hence exponential part decreases throughout the interval, making

$$f(x)$$

strictly decreasing on

$$[-3,0]$$

Therefore, absolute maximum occurs at

$$x=-3$$

Hence,

$$\boxed{\alpha=-3}$$