Given that the coordinates of A is (2, -3), $\overrightarrow{AB}=\begin{pmatrix}-5\\7\end{pmatrix}$ and $\overrightarrow{FE}=k\begin{pmatrix}2.5\\-3.5\end{pmatrix}$.

1. Relate the given vectors using the parallelogram property

In parallelogram $ABFE$, $\overrightarrow{AB}$ and $\overrightarrow{FE}$ are opposite sides but in opposite directions.

So,
$\overrightarrow{FE}=-\overrightarrow{AB}.$

2. Substitute the given vectors and solve for $k$

Given
$\overrightarrow{AB}=\begin{pmatrix}-5\\7\end{pmatrix}, \qquad \overrightarrow{FE}=k\begin{pmatrix}2.5\\-3.5\end{pmatrix}.$

Hence
$k\begin{pmatrix}2.5\\-3.5\end{pmatrix}=-\begin{pmatrix}-5\\7\end{pmatrix}=\begin{pmatrix}5\\-7\end{pmatrix}.$

Comparing components,
$2.5k=5$
$k=2.$

1. Add the displacement vector $\overrightarrow{AB}$ to the coordinates of $A$

Since $A=(2,-3)$ and $\overrightarrow{AB}=\begin{pmatrix}-5\\7\end{pmatrix}$,
$B=(2+(-5),\,-3+7)=(-3,4).$

1. Find the vector from A to C

Given $A=(2,-3)$ and $C=(6,-10)$,
$\overrightarrow{AC}=\begin{pmatrix}6-2\\-10-(-3)\end{pmatrix}=\begin{pmatrix}4\\-7\end{pmatrix}.$

2. Compare $\overrightarrow{AB}$ and $\overrightarrow{AC}$ to justify collinearity

$\overrightarrow{AB}=\begin{pmatrix}-5\\7\end{pmatrix},\qquad \overrightarrow{AC}=\begin{pmatrix}4\\-7\end{pmatrix}.$
For collinearity, one vector must be a scalar multiple of the other.

But
$\frac{-5}{4}\ne \frac{7}{-7}.$
So $\overrightarrow{AB}$ is not parallel to $\overrightarrow{AC}$.