More than enough info!

In $\Delta ABC$ , we know the following identity:

$\displaystyle \sum_{cyc} \cot A = \dfrac{\displaystyle\sum_{cyc} a^2}{4[ABC]} \\ \Rightarrow \sum_{cyc} \cot A = \dfrac{27}{4 \times \dfrac{9\sqrt{3}}{4}} \\ \Rightarrow \sum_{cyc} \cot A = \dfrac{27}{9\sqrt{3}} = \dfrac{3}{\sqrt{3}} \\ \Rightarrow \sum_{cyc} \cot A = \sqrt{3}$

But in $\Delta ABC$ , the following inequality holds : $\displaystyle\sum_{cyc} \cot A \geq \sqrt{3}$ and equality if and only if $\Delta ABC$ is equilateral $\Rightarrow a=b=c$ .

Thus , $a^2+a^2+a^2=27 \Rightarrow 3a^2=27 \Rightarrow a^2=9 \Rightarrow a=b=c=3$

Thus , $\dfrac{abc}{a+b+c} = \dfrac{3\times 3\times 3}{3+3+3} = \dfrac{27}{9} = \Large\boxed{3}$

since 27 is an odd number so $a^2$ , $b^2$ , $c^2$ are odd numbers too

the choices we have are: 1 , 3 , 5

now it is clearly that 3 might be our number

so we have equilateral triangle with side length 3

by using our trig we can figure out the height which equal to $\frac{3\sqrt{3} }{2}$

so 3 is our number and also our answer