Given the vectors OA = (a−1)i + 2j and OB = 3ai + 6j. a) Find the vector AB in terms of a. b) Find the value of a given that AB = 4i + 2j.

The vector $\vec{AB}$ can be found by subtracting vector $\vec{OA}$ from vector $\vec{OB}$:

$$\vec{AB} = \vec{OB} - \vec{OA}$$

$$\vec{AB} = (3a - a + 1)\mathbf{i} + (4)\mathbf{j}$$ $$\vec{AB} = (2a + 1)\mathbf{i} + 4\mathbf{j}$$

The vector $\vec{AB}$ in terms of $a$ is $\boxed{(2a+1)\mathbf{i} + 4\mathbf{j}}$.

We have $\vec{AB} = (2a+1)\mathbf{i} + 4\mathbf{j}$ and we are given $\vec{AB} = 4\mathbf{i} + 2\mathbf{j}$.
Therefore:

$$(2a+1)\mathbf{i} + 4\mathbf{j} = 4\mathbf{i} + 2\mathbf{j}$$

For the $\mathbf{i}$ components:

$$2a+1 = 4$$

For the $\mathbf{j}$ components:

$$4 = 2$$

The equation from the $\mathbf{j}$ components, $4=2$, is a contradiction. This indicates there might be an error in the problem statement as it's impossible for the $\mathbf{j}$ components to be equal.
However, if we assume there's a typo in the given $\vec{AB}$ for part (b) and proceed with the $\mathbf{i}$ components to find $a$:

$$2a+1 = 4$$ $$2a = 4 - 1$$ $$2a = 3$$ $$a = \frac{3}{2}$$

Given the contradiction in the $\mathbf{j}$ components, the problem as stated has no solution for $a$ that satisfies both components simultaneously. If we are forced to choose one component, the $\mathbf{i}$ component gives $a = \frac{3}{2}$.

Assuming the question intended for the $\mathbf{j}$ component of $\vec{AB}$ to be $4\mathbf{j}$ in part (b) as well, then the value of $a$ would be derived solely from the $\mathbf{i}$ components.
The value of $a$ is $\boxed{\frac{3}{2}}$.