Given,

 $\overrightarrow{u}=a\left({\log }_{e}b\right)\hat{i}-6\hat{j}+3\hat{k}$ and $\overrightarrow{v}=\left({\log }_{e}b\right)\hat{i}+2\hat{j}+2a\left({\log }_{e}b\right)\hat{k}$ 

For angle to be acute $\overrightarrow{u}·\overrightarrow{v}>0$

$\Rightarrow \left(a\left({\log }_{e}b\right)\hat{i}-6\hat{j}+3\hat{k}\right).\left(\left({\log }_{e}b\right)\hat{i}+2\hat{j}+2a\left({\log }_{e}b\right)\hat{k}\right)>0$

$\Rightarrow a{\left({\log }_{e}b\right)}^2-12+6a\left({\log }_{e}b\right)>0$ $\forall b>1$

Let $y=a{\left({\log }_{e}b\right)}^2-12+6a\left({\log }_{e}b\right)$

Now let ${\log }_{e}b=t\Rightarrow t>0$ as $b>1$

Now $y=a{t}^2+6at-12>0,\forall t>0$

Since $y>0$, so $a>0 \& D<0$

$\Rightarrow D=36a^2+48a<0$

$\Rightarrow a\in \left(\frac{-4}{3},0\right)$ but $a>0$,

So, $a\in \varphi$