If $$\lim_{x \to 1} \left(\frac{\sin(3x^2 - 4x + 1) - x^2 + 1}{2x^3 - 7x^2 + ax + b}\right) = -2$$, then the value of $$(a - b)$$ is equal to

We need to find $$(a - b)$$ given $$\lim_{x \to 1}\frac{\sin(3x^2 - 4x + 1) - x^2 + 1}{2x^3 - 7x^2 + ax + b} = -2$$.

Check that numerator and denominator both vanish at $$x = 1$$:

Numerator at $$x = 1$$: $$\sin(3 - 4 + 1) - 1 + 1 = \sin(0) = 0$$ $$\checkmark$$

Denominator at $$x = 1$$: $$2 - 7 + a + b = a + b - 5$$

For the limit to exist and be finite, we need $$a + b - 5 = 0$$, i.e., $$a + b = 5 \quad \cdots (1)$$.

Apply L'Hopital's rule:

Numerator derivative: $$\cos(3x^2 - 4x + 1)(6x - 4) - 2x$$

At $$x = 1$$: $$\cos(0)(2) - 2 = 2 - 2 = 0$$

Denominator derivative: $$6x^2 - 14x + a$$

At $$x = 1$$: $$6 - 14 + a = a - 8$$

For 0/0 form again, we need $$a - 8 = 0$$, so $$a = 8$$ and from (1), $$b = -3$$.

Apply L'Hopital's rule again:

Numerator second derivative:

$$\frac{d}{dx}[\cos(3x^2 - 4x + 1)(6x-4) - 2x]$$

$$= -\sin(3x^2 - 4x + 1)(6x-4)^2 + \cos(3x^2 - 4x + 1) \cdot 6 - 2$$

At $$x = 1$$: $$-\sin(0)(4) + \cos(0)(6) - 2 = 0 + 6 - 2 = 4$$

Denominator second derivative: $$12x - 14$$

At $$x = 1$$: $$12 - 14 = -2$$

Compute the limit:

$$\lim_{x \to 1} = \frac{4}{-2} = -2$$ $$\checkmark$$

Find $$a - b$$:

$$a - b = 8 - (-3) = 11$$

The answer is $$\boxed{11}$$.