Relative to the origin $O$, three distinct fixed points $A$, $B$ and $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively. It is known that $\mathbf{b}$ is a unit vector, $|\mathbf{a}| = 3$, $|\mathbf{c}| = 2$ and angle $AOC = 60^\circ$.

1. State the geometrical interpretation using the fact that b is a unit vector

We are given that $\mathbf{b}$ is a unit vector ($|\mathbf{b}| = 1$).
The dot product $\mathbf{b} \cdot \mathbf{c}$ gives the scalar projection of $\mathbf{c}$ onto $\mathbf{b}$.
Taking the modulus $|\mathbf{b} \cdot \mathbf{c}|$ gives the absolute length of this projection.
Thus, it represents the length of projection of $OC$ onto the line $OB$.

1. Express the areas of the two triangles using the vector cross product.

The area of a triangle formed by two position vectors is half the magnitude of their cross product.
$\text{Area of } \triangle AOB = \frac{1}{2} |\mathbf{a} \times \mathbf{b}|$
$\text{Area of } \triangle BOC = \frac{1}{2} |\mathbf{b} \times \mathbf{c}|$

2. Manipulate the given vector equation to determine the ratio of the areas.

Given the vector equation:
$\mathbf{a} \times 2\mathbf{b} = \mathbf{b} \times \mathbf{c}$

By the properties of the cross product, the scalar multiple can be factored out:
$2(\mathbf{a} \times \mathbf{b}) = \mathbf{b} \times \mathbf{c}$

Taking the magnitude on both sides gives:
$2|\mathbf{a} \times \mathbf{b}| = |\mathbf{b} \times \mathbf{c}|$
$\frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{b} \times \mathbf{c}|} = \frac{1}{2}$

Therefore, the ratio of the areas is:
$\frac{\text{Area of } \triangle AOB}{\text{Area of } \triangle BOC} = \frac{\frac{1}{2} |\mathbf{a} \times \mathbf{b}|}{\frac{1}{2} |\mathbf{b} \times \mathbf{c}|} = \frac{1}{2}$

1. Rearrange the given vector product equation to group terms with $\mathbf{b}$.

Given $\mathbf{a} \times 2\mathbf{b} = \mathbf{b} \times \mathbf{c}$, we use the anti-commutative property of vector products ($\mathbf{b} \times \mathbf{c} = -\mathbf{c} \times \mathbf{b}$):
$2(\mathbf{a} \times \mathbf{b}) = -(\mathbf{c} \times \mathbf{b})$
$2(\mathbf{a} \times \mathbf{b}) + (\mathbf{c} \times \mathbf{b}) = \mathbf{0}$
Using distributivity:
$(2\mathbf{a} + \mathbf{c}) \times \mathbf{b} = \mathbf{0}$

2. Deduce the parallel relationship to show the required expression.

Since the cross product of $(2\mathbf{a} + \mathbf{c})$ and $\mathbf{b}$ is the zero vector, the two vectors must be parallel.
Hence, $2\mathbf{a} + \mathbf{c} = k\mathbf{b}$ for some scalar $k \in \mathbb{R}$.
Because $\mathbf{a}$ and $\mathbf{c}$ are non-zero vectors separated by $60^\circ$, they are not collinear, so $2\mathbf{a} + \mathbf{c} \ne \mathbf{0}$. Thus, $k \ne 0$. (Shown)

3. Expand the scalar product $(2\mathbf{a} + \mathbf{c}) \cdot (2\mathbf{a} + \mathbf{c})$.

Taking the scalar product of the expression with itself:
$(2\mathbf{a} + \mathbf{c}) \cdot (2\mathbf{a} + \mathbf{c}) = (k\mathbf{b}) \cdot (k\mathbf{b})$
Expanding the left-hand side:
$4(\mathbf{a} \cdot \mathbf{a}) + 2(\mathbf{a} \cdot \mathbf{c}) + 2(\mathbf{c} \cdot \mathbf{a}) + (\mathbf{c} \cdot \mathbf{c}) = k^2(\mathbf{b} \cdot \mathbf{b})$
$4|\mathbf{a}|^2 + 4(\mathbf{a} \cdot \mathbf{c}) + |\mathbf{c}|^2 = k^2|\mathbf{b}|^2$

4. Substitute the known magnitudes and angle to evaluate the scalar product equation.

Using the definition $\mathbf{a} \cdot \mathbf{c} = |\mathbf{a}||\mathbf{c}|\cos(60^\circ)$ and substituting $|\mathbf{a}| = 3$, $|\mathbf{c}| = 2$, and $|\mathbf{b}| = 1$:
$4(3)^2 + 4(3)(2)\cos(60^\circ) + (2)^2 = k^2(1)^2$
$36 + 24\left(\frac{1}{2}\right) + 4 = k^2$
$36 + 12 + 4 = k^2$
$52 = k^2$

5. Solve for the exact values of $k$.

$k = \pm \sqrt{52}$
$k = \pm 2\sqrt{13}$