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(B) For the angle between two vectors $\vec{u}$ and $\vec{v}$ to be acute,their dot product must be positive,i.e.,$\vec{u} \cdot \vec{v} > 0$.Given $\

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$\phi$

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(B) For the angle between two vectors $\vec{u}$ and $\vec{v}$ to be acute,their dot product must be positive,i.e.,$\vec{u} \cdot \vec{v} > 0$.Given $\vec{u} = a(\log_{e} b) \hat{i} - 6 \hat{j} + 3 \hat{k}$ and $\vec{v} = (\log_{e} b) \hat{i} + 2 \hat{j} + 2a(\log_{e} b) \hat{k}$.Calculating the dot product: $\vec{u} \cdot \vec{v} = a(\log_{e} b)^2 - 12 + 6a(\log_{e} b) > 0$.Let $t = \log_{e} b$. Since $b > 1$,we have $t > 0$.The inequality becomes $at^2 + 6at - 12 > 0$ for all $t > 0$.If $a \le 0$,then for large $t$,$at^2 + 6at - 12$ will be negative,so $a$ must be positive.For a quadratic $f(t) = at^2 + 6at - 12$ to be positive for all $t > 0$,the vertex must be at $t = -\frac{6a}{2a} = -3$. Since the vertex is at $t = -3$ (which is outside the domain $t > 0$) and the parabola opens upward $(a > 0)$,the minimum value for $t > 0$ occurs as $t 	o 0^+$.As $t 	o 0^+$,$f(t) 	o -12$,which is not greater than $0$.Thus,there is no value of $a$ that satisfies the condition for all $t > 0$.Therefore,$S = \phi$.

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 $\vec{v} = (\log_{e} b) \hat{i} + 2 \hat{j} + 2a(\log_{e} b) \hat{k}$ is acute,where $b > 1$. Then $S$ is equal to:

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