Let $n$ be the number of pairs of values of $b$ and $c$ such that $3x+by+c=0$ and $cx-2y+12=0$ have the same graph. Then $n$ is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ \text{finite but more than 2}\qquad \textbf{(E)}\ \infty$
For two lines to be the same, their slopes must be equal and their intercepts must be equal.  This is a necessary and sufficient condition.
The slope of the first line is $\frac{-3}{b}$, while the slope of the second line is $\frac{c}{2}$.  Thus, $\frac{-3}{b} = \frac{c}{2}$, or $bc = -6$.
The intercept of the first line is $\frac{-c}{b}$, while the intercept of the second line is $6$.  Thus, $6 = \frac{-c}{b}$, or $-6b = c$.
Plugging $-6b = c$ into $bc = -6$ gives $b(-6b) = -6$, or $b^2 = 1$.  This means $b = \pm 1$  This in turn gives $c = \mp 6$.  Thus, $(b, c) = (\pm 1, \mp 6)$, for two solutions, which is answer $\boxed{\textbf{(C)}}$