Find the coordinates of the foci and the vertices, the eccentricity,

The given equation is $\frac{y^{2}}{9}-\frac{x^{2}}{27}=1$ or $\frac{y^{2}}{3^{2}}-\frac{x^{2}}{(\sqrt{27})^{2}}=1$.

On comparing this equation with the standard equation of hyperbola i.e., $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$, we obtain $a=3$ and $b=\sqrt{27}$.

We know that $a^{2}+b^{2}=c^{2}$.

$\therefore c^{2}=3^{2}+(\sqrt{27})^{2}=9+27=36$

$\Rightarrow c=6$

The coordinates of the foci are (0, ±6).

The coordinates of the vertices are (0, ±3).

Eccentricity, $e=\frac{c}{a}=\frac{6}{3}=2$

Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 27}{3}=18$