Question about a pair of straight lines

Find the centroid of the triangle formed by the pair of straight lines $12x^2-20xy+7y^2=0$ and the line $2x-3y+4=0$.

My doubt is:

The given pair of straight lines and the third line all pass through the point $(1,2)$. So how can three concurrent straight lines form a triangle? If the question has no flaw, please help me with it.

-

$12x^2−20xy+7y^2=0$
$(6x-7y)(2x-y) = 0$
So these are two lines and $(1,2)$ satisfies only one of them, not both of them . They form a triangle .
So the centroid is $\left(\frac83,\frac83\right)$ – Tejas Dec 5 '13 at 9:43