$2|xy-3y-4x+12| = a^2+2a-z-30$ and $3a^2-a-z-32=0$ and $z-x^2-y^2+6x+8y=0$
What I've got: $3a^2-a-32=x^2+y^2-6x-8y$ so $(x-3)^2 + (y-4)^2 = 3a^2-a-7$ and this is circle
$2|xy-3y-4x+12|=-2a^2+3a+2$. Assume value inside module > 0: $2xy-6y-8x+24=-2a^2+3a+2$, so $(-2a^2+3a+2)/(2(x-3)) + 4 = y$ - it is hyperbola(reflected relatively y-axis). And I stuck.
the term between the absolut value signs can written as $$(x-3)(y-4)$$ now you can do case work