Answer: (a) \(\begin{pmatrix}-5\\12\end{pmatrix}\); (b) \(13\); (c) \(\begin{pmatrix}-\frac{11}{2}\\12\end{pmatrix}\).

Write each position or direction vector in terms of the given basis vectors, then compare coefficients to find the required constants or ratio.

(a) The displacement vector of \(B\) from \(A\) is

\(\overrightarrow{AB}=\begin{pmatrix}-3\\6\end{pmatrix}-\begin{pmatrix}2\\-6\end{pmatrix}\).

So

\(\overrightarrow{AB}=\begin{pmatrix}-5\\12\end{pmatrix}\).

(b) The distance \(AB\) is the magnitude of this vector:

\(AB=\sqrt{(-5)^2+12^2}\).

Thus

\(AB=\sqrt{25+144}=\sqrt{169}=13\).

(c) We are given

\(3\overrightarrow{AB}=2\overrightarrow{AX}\).

Hence

\(\overrightarrow{AX}=\frac32\overrightarrow{AB}\).

So

\(\overrightarrow{AX}=\frac32\begin{pmatrix}-5\\12\end{pmatrix}=\begin{pmatrix}-\frac{15}{2}\\18\end{pmatrix}\).

The position vector of \(X\) is

\(\overrightarrow{OX}=\overrightarrow{OA}+\overrightarrow{AX}\).

Therefore

\(\overrightarrow{OX}=\begin{pmatrix}2\\-6\end{pmatrix}+\begin{pmatrix}-\frac{15}{2}\\18\end{pmatrix}=\begin{pmatrix}-\frac{11}{2}\\12\end{pmatrix}\).