For how many integer values of $m$ parabola $y=(m-2)x^2+12x+m+3$ passes through only three quadrants? 
For how many integer values of $m$ the graph of parabola
$y=(m-2)x^2+12x+m+3$ passes through only three quadrants?
$1)0\qquad\qquad2)7\qquad\qquad3)12\qquad\qquad4)\text{infinity}$

If the parabola $ax^2+bx+c=0$ passes through three quadrant I think we should have $\frac ca>0$ and $\Delta>0$:
$$\frac{m+3}{m-2}>0\Rightarrow m\in(-\infty,-3] \cup(2,+\infty)$$
$$\Delta'>0\Rightarrow 36-(m-2)(m+3)>0\Rightarrow m^2+m-42<0\Rightarrow m\in [-7,6]$$
Also for $m=2$ we have $y=12x+5$ and it passes through three quadrants.
So $m$ can be $-7,-6,\cdots,-3, $ or $2,3,4,5,6$ . so there are $10$ possible values for $m$ but this isn't in the options. What am I missing?

I was just doing the same observation as @Arctic Char.
Here is a graphical representation confirming that (the values of $m$ are superimposed on the curves):

