In the parallelogram OABC, OA = 4q, OC = 6c, AP = 2/3 AC and M is the midpoint of BC. Find in terms of q and/or c: (a) AC (b) OP (c) OM. Hence or otherwise, show that the points O, P and M are collinear.

The vector $\vec{AC}$ is $\boxed{-4q + 6c}$.

The vector $\vec{OP}$ is $\boxed{\frac{4}{3}q + 4c}$.

In parallelogram OABC, $\vec{CB} = \vec{OA}$. Since M is the midpoint of BC, $\vec{CM} = \frac{1}{2}\vec{CB}$.
Therefore, $\vec{CM} = \frac{1}{2}\vec{OA}$.

The vector $\vec{OM}$ is $\boxed{2q + 6c}$.

From part (i), we have:

Factor out $\frac{4}{3}$ from $\vec{OP}$:

Factor out 2 from $\vec{OM}$:

We can see that $(q + 3c)$ is a common vector.
From $\vec{OM} = 2(q + 3c)$, we can write $(q + 3c) = \frac{1}{2}\vec{OM}$.
Substitute this into the expression for $\vec{OP}$:

Since $\vec{OP} = \frac{2}{3}\vec{OM}$, $\vec{OP}$ is a scalar multiple of $\vec{OM}$. Both vectors originate from the common point O. Therefore, the points O, P, and M are collinear.