With reference to the origin \(O\), the points \(A\), \(B\) and \(C\) are such that vectors \(\overrightarrow{OA}=\mathbf{a}\), \(\overrightarrow{OB}=\mathbf{b}\) and \(\overrightarrow{OC}=2\mathbf{a}-2\mathbf{b}\).
Given that \(|\mathbf{a}|=2\), \(|\mathbf{b}|=3\) and the angle between \(\mathbf{a}\) and \(\mathbf{b}\) is \(\frac{\pi}{6}\), find the area of triangle \(OBC\).[3]
The lines \(OA\) and \(BC\) intersect at point \(X\).
Show that \(\displaystyle \overrightarrow{OX}=\frac23\mathbf{a}\).[3]
The plane \(p\) has equation \(\mathbf{r}\cdot\mathbf{n}=d\) and contains the point \(A\), where \(\mathbf{n}\) is a unit vector.
Find the shortest distance from point \(X\) to plane \(p\), in the form of \(\frac1k|d|\), where \(k\) is a positive integer to be determined.[3]