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. Rearrange the given equation to collect terms involving the cross product with $\mathbf{b}$.

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

2. Interpret the meaning of the cross product being the zero vector.

Since it is given that $3\mathbf{a} \ne -2\mathbf{c}$, we have $3\mathbf{a} + 2\mathbf{c} \ne \mathbf{0}$.
It is also given that $\mathbf{b}$ is a non-zero vector.
For the cross product of two non-zero vectors to be $\mathbf{0}$, the vectors must be parallel to each other.

3. State the linear relationship between the vectors.

Since $3\mathbf{a} + 2\mathbf{c}$ is parallel to $\mathbf{b}$, we can express one as a scalar multiple of the other:
$3\mathbf{a} + 2\mathbf{c} = \lambda\mathbf{b}$
where $\lambda \in \mathbb{R}$. Furthermore, since $3\mathbf{a} + 2\mathbf{c} \ne \mathbf{0}$, $\lambda$ must be non-zero.

1. State the condition that the vector cross product must be perpendicular to its operands.

The vector product $\mathbf{v}\times(-\mathbf{i}+2\mathbf{j}+\mathbf{k})$ yields a vector that is perpendicular to both $\mathbf{v}$ and $(-\mathbf{i}+2\mathbf{j}+\mathbf{k})$.
Therefore, the scalar product of the result $(a\mathbf{i} + \mathbf{j} + \mathbf{k})$ and $(-\mathbf{i}+2\mathbf{j}+\mathbf{k})$ must be zero:
$\begin{pmatrix} a \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} = 0$

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

$a(-1) + 1(2) + 1(1) = 0$
$-a + 3 = 0$
$a = 3$

1. Express $\mathbf{v}$ in Cartesian components and expand the cross product equation.

Let $\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$.
Using the result $a=3$ from part (b)(i), the equation becomes:
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \times \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}$
$\begin{pmatrix} y(1) - z(2) \\ z(-1) - x(1) \\ x(2) - y(-1) \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}$
$\begin{pmatrix} y - 2z \\ -x - z \\ 2x + y \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}$

2. Solve the system of linear equations to find the general set of position vectors $\mathbf{v}$.

Equating the components gives a system of equations:
$y - 2z = 3 \implies y = 3 + 2z$
$-x - z = 1 \implies x = -1 - z$
$2x + y = 1$
Let $z = \lambda$ where $\lambda \in \mathbb{R}$.
Then $x = -1 - \lambda$ and $y = 3 + 2\lambda$.
(Checking with the third equation: $2(-1-\lambda) + 3+2\lambda = 1$, which is consistent).

Hence, $\mathbf{v} = \begin{pmatrix} -1-\lambda \\ 3+2\lambda \\ \lambda \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}$.

The set of position vectors is $\{ \mathbf{v} = -\mathbf{i} + 3\mathbf{j} + \lambda(-\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \mid \lambda \in \mathbb{R} \}$.

3. Describe the set of points geometrically.

The equation $\mathbf{v} = (-\mathbf{i} + 3\mathbf{j}) + \lambda(-\mathbf{i} + 2\mathbf{j} + \mathbf{k})$ is in the standard form of the vector equation of a line $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$.

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