Answer: 136

Small Hint:

The sum of the roots, $4w + 12i - 4,$ must be real; this determines the imaginary part of $w$

Big Hint:

Real coefficients force the two non-real roots to be conjugates, and $1 + a + b + c = P(1)$

Solution:

Write $w = x + yi$ with $x$ and $y$ real. The sum of the roots is $4w + 12i - 4 = -a,$ which is real, so $4y + 12 = 0$ and $y = -3.$ The roots are then $x,$ $x + 6i,$ and $2x - 4 - 6i.$ Because the coefficients are real, the two non-real roots must be conjugates, so $2x - 4 = x,$ giving $x = 4.$ The roots are $4,$ $4 + 6i,$ and $4 - 6i.$

Now $$\begin{aligned} 1 + a + b + c &= P(1) \\ &= (1 - 4) \\ &\quad {}\cdot \bigl(1 - (4 + 6i)\bigr) \\ &\quad {}\cdot \bigl(1 - (4 - 6i)\bigr) \\ &= (-3)(9 + 36) \\ &= -135, \end{aligned}$$ so $a + b + c = -136$ and $|a + b + c| = 136.$