Let $$\vec{a} = \alpha \hat{i} + \hat{j} + \beta \hat{k}$$ and $$\vec{b} = 3\hat{i} - 5\hat{j} + 4\hat{k}$$ be two vectors, such that $$\vec{a} \times \vec{b} = -\hat{i} + 9\hat{j} + 12\hat{k}$$. Then the projection of $$\vec{b} - 2\vec{a}$$ on $$\vec{b} + \vec{a}$$ is equal to

For $$x \leq 1$$: $$f(x) = x^3 - x^2 + 10x - 7$$, so $$f(1) = 1 - 1 + 10 - 7 = 3$$.

$$f'(x) = 3x^2 - 2x + 10$$. The discriminant is $$4 - 120 = -116 < 0$$, and the leading coefficient is positive, so $$f'(x) > 0$$ for all $$x$$. This means $$f$$ is strictly increasing on $$(-\infty, 1]$$, and $$f(1) = 3$$ is the maximum on this interval.

For $$x > 1$$: $$f(x) = -2x + \log_2(b^2 - 4)$$. Since the coefficient of $$x$$ is $$-2 < 0$$, this is a decreasing function. Its supremum as $$x \to 1^+$$ is $$-2 + \log_2(b^2 - 4)$$.

For $$f(x)$$ to have maximum at $$x = 1$$, we need:

(i) $$\log_2(b^2-4)$$ must be defined: $$b^2 - 4 > 0$$, i.e., $$|b| > 2$$, so $$b \in (-\infty, -2) \cup (2, \infty)$$.

(ii) $$f(1^+) \leq f(1)$$: $$-2 + \log_2(b^2 - 4) \leq 3$$, i.e., $$\log_2(b^2 - 4) \leq 5$$.

$$b^2 - 4 \leq 2^5 = 32$$, so $$b^2 \leq 36$$, giving $$|b| \leq 6$$, i.e., $$b \in [-6, 6]$$.

Taking the intersection of (i) and (ii): $$b \in [-6, -2) \cup (2, 6]$$.

The answer is Option C: $$[-6, -2) \cup (2, 6]$$.