Given points A(-2, 3), B(-6, 1), C(12, -1), and D(4, -5), show that vectors AB and CD are parallel. Also, find vector AB where A is the midpoint of M(5, 6) and N(-1, 4).

Given $A(-2, 3)$ and $B(-6, 1)$.

Given $C(12, -1)$ and $D(4, -5)$.

We observe that $\vec{CD} = \begin{pmatrix} -8 \\ -4 \end{pmatrix} = 2 \begin{pmatrix} -4 \\ -2 \end{pmatrix} = 2 \vec{AB}$.
Since $\vec{CD}$ is a scalar multiple of $\vec{AB}$ (with scalar $k=2$), the vectors are parallel.
Thus, $\vec{AB}$ and $\vec{CD}$ are indeed parallel.

Given $M(5, 6)$ and $N(-1, 4)$. The midpoint formula is $A = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)$.

Given $B(-1, 5)$ and $A(2, 5)$.

The vector $\vec{AB}$ is $\boxed{\begin{pmatrix} \textbf{-3} \\ \textbf{0} \end{pmatrix}}$.

Add $\begin{pmatrix} 3 \\ 7 \end{pmatrix}$ to both sides:

Divide both sides by 2:

The vector $f$ is $\boxed{\begin{pmatrix} \textbf{2} \\ \textbf{5} \end{pmatrix}}$.