Relative to the origin $O$, two points $A$ and $B$ have position vectors given by $\mathbf{a} = 2\mathbf{i} - \mathbf{j}$ and $\mathbf{b} = \alpha \mathbf{k}$ respectively, where $\alpha$ is a negative real constant.

1. Evaluate the scalar product of the position vectors $\mathbf{a}$ and $\mathbf{b}$

$\mathbf{a} = \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 0 \\ 0 \\ \alpha \end{pmatrix}$

$\mathbf{a} \cdot \mathbf{b} = (2)(0) + (-1)(0) + (0)(\alpha) = 0$

2. Deduce the angle between the two vectors

Since $\mathbf{a} \cdot \mathbf{b} = 0$ and both vectors are non-zero, $OA$ is perpendicular to $OB$.

Therefore, the angle is $90^\circ$.

1. Find the two possible values of $\theta$ from the given sine value.

Given $\sin \theta = \frac{1}{2}$ and knowing that $\theta$ is an interior angle of $\triangle OAB$ ($0^\circ < \theta < 180^\circ$):
$\theta = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ$
$\text{or } \theta = 180^\circ - 30^\circ = 150^\circ$

2. Justify the correct value using the angle between $OA$ and $OB$.

To determine the angle between $OA$ and $OB$, we evaluate the scalar product of their position vectors (which serves as a required result from part a):
$\mathbf{a} \cdot \mathbf{b} = \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ \alpha \end{pmatrix} = (2)(0) + (-1)(0) + (0)(\alpha) = 0$
Since the scalar product is $0$, $\angle AOB = 90^\circ$.
As $\triangle OAB$ is a right-angled triangle, the other two angles (including $\angle OBA = \theta$) must be acute ($\theta < 90^\circ$).
Therefore, the correct value is $\theta = 30^\circ$.

1. Set up the scalar product equation using vectors $\vec{BO}$ and $\vec{BA}$ and the known angle from part (b).

From part (b), we know $\angle OBA = 30^\circ$.
Using the geometric definition of the scalar product for vectors originating from $B$:
$\vec{BO} \cdot \vec{BA} = |\vec{BO}| |\vec{BA}| \cos(30^\circ)$

2. Evaluate the components, magnitudes, and algebraic scalar product of $\vec{BO}$ and $\vec{BA}$.

$\vec{BO} = -\mathbf{b} = -\alpha \mathbf{k}$
$\vec{BA} = \mathbf{a} - \mathbf{b} = 2\mathbf{i} - \mathbf{j} - \alpha \mathbf{k}$
Evaluating the scalar product algebraically:
$\vec{BO} \cdot \vec{BA} = (0)(2) + (0)(-1) + (-\alpha)(-\alpha) = \alpha^2$
Evaluating the magnitudes:
$|\vec{BO}| = \sqrt{(-\alpha)^2} = \sqrt{\alpha^2}$
$|\vec{BA}| = \sqrt{2^2 + (-1)^2 + (-\alpha)^2} = \sqrt{5 + \alpha^2}$
Substituting these into the equation from Step 1:
$\alpha^2 = \sqrt{\alpha^2} \sqrt{5 + \alpha^2} \left(\frac{\sqrt{3}}{2}\right)$

3. Solve the resulting equation for the exact value of $\alpha$, taking into account that $\alpha < 0$.

Since $\alpha < 0$, we have $\sqrt{\alpha^2} = |\alpha| = -\alpha$. Substituting this into the equation gives:
$\alpha^2 = (-\alpha) \sqrt{5 + \alpha^2} \left(\frac{\sqrt{3}}{2}\right)$
Dividing both sides by $-\alpha$ (which is valid since $\alpha \neq 0$):
$-\alpha = \frac{\sqrt{3}}{2}\sqrt{5 + \alpha^2}$
Squaring both sides:
$\alpha^2 = \frac{3}{4}(5 + \alpha^2)$
$4\alpha^2 = 15 + 3\alpha^2$
$\alpha^2 = 15$
$\alpha = \pm\sqrt{15}$
Since $\alpha$ is given as a negative constant:
$\alpha = -\sqrt{15}$

1. Identify and justify the shape of quadrilateral $OABC$.

The position vector of $C$ is $\mathbf{c} = \mathbf{b} - \mathbf{a}$, which means the displacement vector $\vec{OC} = \mathbf{b} - \mathbf{a}$.
The displacement vector $\vec{AB}$ is given by $\mathbf{b} - \mathbf{a}$.
Since $\vec{OC} = \vec{AB}$, the line segments $OC$ and $AB$ are parallel and equal in length.
Therefore, $OABC$ is a parallelogram.

2. Find the magnitudes of vectors $\mathbf{a}$ and $\mathbf{b}$ using prior parts.

First, we calculate the magnitude of $\mathbf{a}$:
$|\mathbf{a}| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{5}$
From part (a), we know $\mathbf{a} \cdot \mathbf{b} = (2)(0) + (-1)(0) + (0)(\alpha) = 0$, which implies that $\triangle OAB$ is a right-angled triangle with $\angle AOB = 90^\circ$.
Using the correct value of $\angle OBA = 30^\circ$ from part (b), we can deduce $|\mathbf{b}|$ (which also confirms the result of part c):
$\tan 30^\circ = \frac{|\mathbf{a}|}{|\mathbf{b}|} \implies |\mathbf{b}| = \frac{\sqrt{5}}{1/\sqrt{3}} = \sqrt{15}$

3. Calculate the exact area of the parallelogram.

Since $OABC$ is a parallelogram, its diagonal $OB$ divides it into two congruent triangles. Its total area is twice the area of the right-angled triangle $\triangle OAB$.
$\text{Area of } OABC = 2 \times \text{Area of } \triangle OAB$
$\text{Area} = 2 \times \left( \frac{1}{2} |\mathbf{a}| |\mathbf{b}| \right)$
$\text{Area} = \sqrt{5} \times \sqrt{15} = \sqrt{75} = 5\sqrt{3}$