If the parabola $y = ax^2 + bx + c$ passes through the points $( - 1, 12)$, $(0, 5)$, and $(2, - 3)$, the value of $a + b + c$ is: 
$\textbf{(A)}\ -4\qquad\textbf{(B)}\ -2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ 2$
Substituting in the $(x, y)$ pairs gives the following system of equations:
\[a-b+c=12\]
\[c=5\]
\[4a+2b+c=-3\]
We know $c=5$, so plugging this in reduces the system to two variables:
\[a-b=7\]
\[4a+2b=-8\]
Dividing the second equation by 2 gives $2a+b=-4$, which can be added to the first equation to get $3a=3$, or $a=1$. So the solution set is $(1, -6, 5)$, and the sum is $\boxed{0\textbf{ (C)}}$.