Step 1: Factor the pair of lines.
\(18x^2-9xy+y^2\) factors neatly as \((6x-y)(3x-y)\), since expanding gives \(18x^2-6xy-3xy+y^2=18x^2-9xy+y^2\). So the two lines through the origin are \(y=6x\) and \(y=3x\).
Step 2: Find where each line meets y = c.
On \(y=6x\): \(x=\frac{c}{6}\). On \(y=3x\): \(x=\frac{c}{3}\). So the triangle has vertices \((0,0)\), \(\left(\frac{c}{6},c\right)\), \(\left(\frac{c}{3},c\right)\).
Step 3: Compute the area.
The side joining \(\left(\frac{c}{6},c\right)\) and \(\left(\frac{c}{3},c\right)\) is horizontal with length \(\frac{c}{3}-\frac{c}{6}=\frac{c}{6}\), and the perpendicular height from \((0,0)\) down to the line \(y=c\) is \(c\). So
\[\text{Area} = \frac{1}{2}\cdot\frac{c}{6}\cdot c = \frac{c^2}{12}\]
Setting this to 27: \(\frac{c^2}{12}=27 \implies c^2 = 324 \implies c = 18\).
Step 4: Find the centroid.
\[G = \left(\frac{0+\frac{c}{6}+\frac{c}{3}}{3}, \frac{0+c+c}{3}\right) = \left(\frac{c}{6}, \frac{2c}{3}\right) = \left(\frac{18}{6}, \frac{36}{3}\right) = (3,12)\]
\[\boxed{(3,12)}\]