Referred to the origin $O$, points $A$, $B$ and $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively. It is given that $\mathbf{a} + \mathbf{b} - \mathbf{c}$ and $\mathbf{a} - \mathbf{b} + \mathbf{c}$ are parallel.

1. Express the condition for the two given vectors to be parallel using the vector product.

$(\mathbf{a} + \mathbf{b} - \mathbf{c}) \times (\mathbf{a} - \mathbf{b} + \mathbf{c}) = \mathbf{0}$

2. Expand the vector product distributively.

$\mathbf{a} \times \mathbf{a} - \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c} + \mathbf{b} \times \mathbf{a} - \mathbf{b} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} - \mathbf{c} \times \mathbf{a} + \mathbf{c} \times \mathbf{b} - \mathbf{c} \times \mathbf{c} = \mathbf{0}$

3. Simplify the expression using the properties $\mathbf{v} \times \mathbf{v} = \mathbf{0}$ and $\mathbf{v} \times \mathbf{w} = -\mathbf{w} \times \mathbf{v}$.

$\mathbf{0} - \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c} - \mathbf{a} \times \mathbf{b} - \mathbf{0} + \mathbf{b} \times \mathbf{c} + \mathbf{a} \times \mathbf{c} - \mathbf{b} \times \mathbf{c} - \mathbf{0} = \mathbf{0}$
$-2(\mathbf{a} \times \mathbf{b}) + 2(\mathbf{a} \times \mathbf{c}) = \mathbf{0}$

4. Factorise the simplified expression to relate it to $\overrightarrow{OA}$ and $\overrightarrow{BC}$, and conclude the proof.

$2\mathbf{a} \times \mathbf{c} - 2\mathbf{a} \times \mathbf{b} = \mathbf{0}$
$2\mathbf{a} \times (\mathbf{c} - \mathbf{b}) = \mathbf{0}$
$\mathbf{a} \times (\mathbf{c} - \mathbf{b}) = \mathbf{0}$
Since $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{BC} = \mathbf{c} - \mathbf{b}$, we have:
$\overrightarrow{OA} \times \overrightarrow{BC} = \mathbf{0}$
Since their vector product is the zero vector, $\overrightarrow{OA}$ and $\overrightarrow{BC}$ are parallel.

1. Find the scalar product $\mathbf{a} \cdot \mathbf{b}$ using the given angle $OBC$.

The angle $OBC$ is the angle between vectors originating from $B$, which are $\overrightarrow{BO}$ and $\overrightarrow{BC}$.
Given $\overrightarrow{BO} = -\mathbf{b}$ and $\overrightarrow{BC} = 2\overrightarrow{OA} = 2\mathbf{a}$.
Using the geometric definition of the scalar product:
$\overrightarrow{BO} \cdot \overrightarrow{BC} = |\overrightarrow{BO}||\overrightarrow{BC}| \cos 60^\circ$
$(-\mathbf{b}) \cdot (2\mathbf{a}) = |-\mathbf{b}| |2\mathbf{a}| \cos 60^\circ$
Since $\mathbf{a}$ and $\mathbf{b}$ are unit vectors, $|-\mathbf{b}| = 1$ and $|2\mathbf{a}| = 2$.
$-2 (\mathbf{a} \cdot \mathbf{b}) = (1)(2)\left(\frac{1}{2}\right)$
$-2 (\mathbf{a} \cdot \mathbf{b}) = 1 \implies \mathbf{a} \cdot \mathbf{b} = -\frac{1}{2}$

2. Express the scalar product $\overrightarrow{OB} \cdot \overrightarrow{OC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

First, express $\mathbf{c}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
From $\overrightarrow{BC} = 2\mathbf{a}$, we have:
$\mathbf{c} - \mathbf{b} = 2\mathbf{a} \implies \mathbf{c} = 2\mathbf{a} + \mathbf{b}$
Now, expand the scalar product $\overrightarrow{OB} \cdot \overrightarrow{OC}$:
$\overrightarrow{OB} \cdot \overrightarrow{OC} = \mathbf{b} \cdot \mathbf{c}$
$= \mathbf{b} \cdot (2\mathbf{a} + \mathbf{b})$
$= 2(\mathbf{a} \cdot \mathbf{b}) + \mathbf{b} \cdot \mathbf{b}$
$= 2(\mathbf{a} \cdot \mathbf{b}) + |\mathbf{b}|^2$

3. Evaluate the expanded scalar product to establish perpendicularity.

Substitute $\mathbf{a} \cdot \mathbf{b} = -\frac{1}{2}$ and $|\mathbf{b}| = 1$ into the expression:
$\overrightarrow{OB} \cdot \overrightarrow{OC} = 2\left(-\frac{1}{2}\right) + 1^2$
$= -1 + 1 = 0$
Since the scalar product is $0$, the vectors $\overrightarrow{OB}$ and $\overrightarrow{OC}$ are perpendicular.

1. Calculate the area of the base $OBC$.

From part (b), we established that $\overrightarrow{OB} \perp \overrightarrow{OC}$, meaning the angle between $\mathbf{b}$ and $\mathbf{c}$ is $90^\circ$. We also know $\mathbf{b}$ is a unit vector ($|\mathbf{b}| = 1$).
The area of the triangular base $OBC$ is given by:
$\text{Base Area} = \frac{1}{2} |\mathbf{b} \times \mathbf{c}| = \frac{1}{2} |\mathbf{b}| |\mathbf{c}| \sin 90^\circ$
Given $|\mathbf{c}| = \sqrt{3}$,
$\text{Base Area} = \frac{1}{2} (1)(\sqrt{3})(1) = \frac{\sqrt{3}}{2}$

2. Determine the perpendicular height of the pyramid.

The vector $\overrightarrow{OH}$ is given by $\mathbf{h} = \mathbf{b} \times \mathbf{c}$.
By the definition of the cross product, $\mathbf{h}$ is perpendicular to both $\mathbf{b}$ and $\mathbf{c}$, which means $\overrightarrow{OH}$ is normal (perpendicular) to the plane containing the base $OBC$.
Therefore, the height of the pyramid is simply the magnitude of $\mathbf{h}$:
$\text{Height} = |\mathbf{h}| = |\mathbf{b} \times \mathbf{c}| = \sqrt{3}$

3. Calculate the final volume of the pyramid.

Using the given formula for the volume of a pyramid:
$\text{Volume} = \frac{1}{3} \times \text{Base Area} \times \text{Height}$
$\text{Volume} = \frac{1}{3} \times \frac{\sqrt{3}}{2} \times \sqrt{3}$
$\text{Volume} = \frac{3}{6} = \frac{1}{2}$