Answer: \(\dfrac{1}{\sqrt{234}}\begin{pmatrix}15\\-3\end{pmatrix}\).

Using position vectors,

\(\overrightarrow{AC}=\begin{pmatrix}x+5\\y+7\end{pmatrix}\)

and

\(\overrightarrow{BC}=\begin{pmatrix}x-10\\y+4\end{pmatrix}.\)

Since \(\overrightarrow{AC}=4\overrightarrow{BC}\),

\(\begin{pmatrix}x+5\\y+7\end{pmatrix} = 4\begin{pmatrix}x-10\\y+4\end{pmatrix}.\)

Equating components gives

\(x+5=4x-40,\)

so \(x=15\), and

\(y+7=4y+16,\)

so \(y=-3\).

Therefore

\(\overrightarrow{OC}=\begin{pmatrix}15\\-3\end{pmatrix}.\)

Its magnitude is

\(\sqrt{15^2+(-3)^2}=\sqrt{234}.\)

Hence a unit vector in the direction of \(\overrightarrow{OC}\) is

\(\boxed{\frac{1}{\sqrt{234}}\begin{pmatrix}15\\-3\end{pmatrix}}.\)