The non-zero vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ are such that $3\mathbf{a} \times \mathbf{b} = 2\mathbf{b} \times \mathbf{c}$, where $3\mathbf{a} \ne -2\mathbf{c}$. Find a linear relationship between $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$. The variable position vector $\mathbf{v}$ satisfies the equation $\mathbf{v} \times (-\mathbf{i} + 2\mathbf{j} + \mathbf{k}) = a\mathbf{i} + \mathbf{j} + \mathbf{k}$, where $a$ is a non-zero constant. Show that $a = 3$. Hence, by finding the set of position vectors $\mathbf{v}$, describe geometrically the set of points represented by the position vectors $\mathbf{v}$.

1. Express the cross products such that they have a common second vector by using anti-commutativity.

$3\mathbf{a} \times \mathbf{b} = 2\mathbf{b} \times \mathbf{c}$
$3\mathbf{a} \times \mathbf{b} = -2\mathbf{c} \times \mathbf{b}$

2. Rearrange and use the distributive property of the cross product to factorise the expression.

$3\mathbf{a} \times \mathbf{b} + 2\mathbf{c} \times \mathbf{b} = \mathbf{0}$
$(3\mathbf{a} + 2\mathbf{c}) \times \mathbf{b} = \mathbf{0}$

3. Interpret the zero cross product to deduce collinearity and formulate a linear relationship.

Given that $3\mathbf{a} \ne -2\mathbf{c}$, the vector $(3\mathbf{a} + 2\mathbf{c})$ is non-zero. Since $\mathbf{b}$ is also non-zero and their cross product is the zero vector, they must be parallel.
$3\mathbf{a} + 2\mathbf{c} = \lambda \mathbf{b}$
where $\lambda$ is a scalar.

1. Establish the orthogonality condition from the vector product equation

The vector product of two vectors is orthogonal (perpendicular) to both original vectors. Therefore, the result $\mathbf{w} = a\mathbf{i} + \mathbf{j} + \mathbf{k}$ must be perpendicular to the vector $\mathbf{n} = -\mathbf{i} + 2\mathbf{j} + \mathbf{k}$.

This means their scalar product must be zero:
$(a\mathbf{i} + \mathbf{j} + \mathbf{k}) \cdot (-\mathbf{i} + 2\mathbf{j} + \mathbf{k}) = 0$

2. Evaluate the scalar product to solve for $a$

$(a)(-1) + (1)(2) + (1)(1) = 0$
$-a + 2 + 1 = 0$
$3 - a = 0$
$a = 3 \text{ (shown)}$

1. Express the given vector equation in terms of Cartesian components to form a system of equations.

Let $\mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$. Using the result $a=3$ from part (i), the equation becomes:
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \times \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}$
Expanding the cross product on the left-hand side:
$\begin{pmatrix} y(1) - z(2) \\ z(-1) - x(1) \\ x(2) - y(-1) \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}$
$\begin{pmatrix} y - 2z \\ -x - z \\ 2x + y \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}$

2. Solve the system of linear equations by expressing $x$ and $y$ in terms of $z$.

Equating the components gives a system of linear equations:

1. $y - 2z = 3 \implies y = 2z + 3$
2. $-x - z = 1 \implies x = -z - 1$
   Substitute these expressions into the third equation to verify consistency:
   $2(-z - 1) + (2z + 3) = -2z - 2 + 2z + 3 = 1$ (consistent).
   Thus, the components are parameterized by $z$: $x = -z - 1$ and $y = 2z + 3$.

3. Express the set of position vectors $\mathbf{v}$ as a vector equation of a line.

Letting $z = \lambda$ where $\lambda \in \mathbb{R}$, we can write the position vector $\mathbf{v}$ as:
$\mathbf{v} = \begin{pmatrix} -\lambda - 1 \\ 2\lambda + 3 \\ \lambda \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}$
In $\mathbf{i}, \mathbf{j}, \mathbf{k}$ notation:
$\mathbf{v} = -\mathbf{i} + 3\mathbf{j} + \lambda(-\mathbf{i} + 2\mathbf{j} + \mathbf{k})$

4. Describe the set of points geometrically.

The set of points represents a line passing through the point with position vector $-\mathbf{i} + 3\mathbf{j}$ (or coordinates $(-1, 3, 0)$) and parallel to the vector $-\mathbf{i} + 2\mathbf{j} + \mathbf{k}$.