Let $f(x)$ = $ax^3+bx^2+cx+d$. Since $f(x)$ is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up. By drawing a coordinate axis, and two lines representing 12 and -12, it is easy to see that f(1)=f(5)=f(6), and f(2)=f(3)=f(7); otherwise more bends would be required in the graph. Since only the absolute value of f(0) is required, there is no loss of generalization by stating that f(1)=12, and f(2)=-12. This provides the following system of equations. \[a + b + c + d = 12\]\[8a + 4b + 2c + d = -12\]\[27a + 9b + 3c + d = -12\]\[125a + 25b + 5c + d = 12\]\[216a + 36b + 6c + d = 12\]\[343a + 49b + 7c + d = -12\] Using any four of these functions as a system of equations yields $|f(0)| = \boxed{072}$