Let S be the set of all $$a \in \mathbb{R}$$ for which the angle between the vectors $$\vec{u} = a(\log_e b)\hat{i} - 6\hat{j} + 3\hat{k}$$ and $$\vec{v} = (\log_e b)\hat{i} + 2\hat{j} + 2a(\log_e b)\hat{k}$$, $$(b > 1)$$ is acute. Then S is equal to

We need the angle between $$\vec{u} = a(\log_e b)\hat{i} - 6\hat{j} + 3\hat{k}$$ and $$\vec{v} = (\log_e b)\hat{i} + 2\hat{j} + 2a(\log_e b)\hat{k}$$ to be acute, where $$b > 1$$.

For the angle to be acute, we need $$\vec{u} \cdot \vec{v} > 0$$. Computing the dot product:

$$\vec{u} \cdot \vec{v} = a(\log_e b)^2 - 12 + 6a(\log_e b)$$

Let $$t = \log_e b$$. Since $$b > 1$$, we have $$t > 0$$. So we need:

$$at^2 + 6at - 12 > 0$$

for ALL $$t > 0$$ (since the condition must hold for every valid $$b > 1$$, meaning every $$t > 0$$).

Let $$f(t) = at^2 + 6at - 12$$. We need $$f(t) > 0$$ for all $$t > 0$$.

Case 1: If $$a = 0$$, then $$f(t) = -12 < 0$$. Not valid.

Case 2: If $$a > 0$$, the parabola opens upward. At $$t = 0$$: $$f(0) = -12 < 0$$. Since the function is continuous and negative at $$t = 0$$, by continuity $$f(t) < 0$$ for small positive $$t$$. So $$f(t) > 0$$ for ALL $$t > 0$$ fails.

Case 3: If $$a < 0$$, the parabola opens downward. As $$t \to \infty$$, $$f(t) \to -\infty$$. So $$f(t) > 0$$ for all $$t > 0$$ is impossible here too (since eventually $$f$$ becomes negative).

So there is no value of $$a$$ for which the angle is acute for all $$b > 1$$.

Hence, $$S = \emptyset$$, and the correct answer is Option B.