Let $P(z)=z^3+az^2+bz+c$, where a, b, and c are real. There exists a complex number $w$ such that the three roots of $P(z)$ are $w+3i$, $w+9i$, and $2w-4$, where $i^2=-1$. Find $|a+b+c|$.
Set $w=x+yi$, so $x_1 = x+(y+3)i$, $x_2 = x+(y+9)i$, $x_3 = 2x-4+2yi$.
Since $a,b,c\in{R}$, the imaginary part of $a,b,c$ must be $0$.
Start with a, since it's the easiest one to do: $y+3+y+9+2y=0, y=-3$,
and therefore: $x_1 = x$, $x_2 = x+6i$, $x_3 = 2x-4-6i$.
Now, do the part where the imaginary part of c is 0 since it's the second easiest one to do: 
$x(x+6i)(2x-4-6i)$. The imaginary part is $6x^2-24x$, which is 0, and therefore $x=4$, since $x=0$ doesn't work.
So now, $x_1 = 4, x_2 = 4+6i, x_3 = 4-6i$,
and therefore: $a=-12, b=84, c=-208$. Finally, we have $|a+b+c|=|-12+84-208|=\boxed{136}$.

Solution 1b
Same as solution 1 except that when you get to $x_1 = x$, $x_2 = x+6i$, $x_3 = 2x-4-6i$, you don't need to find the imaginary part of $c$. We know that $x_1$ is a real number, which means that $x_2$ and $x_3$ are complex conjugates. Therefore, $x=2x-4$.