Three non-zero vectors $\mathbf{p}$, $\mathbf{q}$ and $\mathbf{r}$ are such that $3\mathbf{p} \times \mathbf{q} = \mathbf{r} \times \mathbf{p}$. Find a linear relationship between $\mathbf{p}$, $\mathbf{q}$ and $\mathbf{r}$. Referred to the origin $O$, the points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$. It is given that $\mathbf{a}$ and $\mathbf{b}$ are non-parallel unit vectors. The point $B$ divides $AC$ in the ratio $1:3$. The point $D$ lies on $OB$ produced such that $OB : OD = 1 : m$, where $m \in \mathbb{Z}$, $m \ge 2$. Given that $\overrightarrow{AB}$ is perpendicular to $\overrightarrow{CD}$, find the numerical value of $m$.

1. Rearrange the given equation and apply the anti-commutative property of the vector product.

$3\mathbf{p} \times \mathbf{q} - \mathbf{r} \times \mathbf{p} = \mathbf{0}$
$3\mathbf{p} \times \mathbf{q} - (-\mathbf{p} \times \mathbf{r}) = \mathbf{0}$

2. Factorise the expression using the distributive property of the vector product.

$3\mathbf{p} \times \mathbf{q} + \mathbf{p} \times \mathbf{r} = \mathbf{0}$
$\mathbf{p} \times (3\mathbf{q} + \mathbf{r}) = \mathbf{0}$

3. Deduce the geometrical relationship between the vectors to form a linear relationship.

Since $\mathbf{p}$ is a non-zero vector, the zero cross product implies that the vector $3\mathbf{q} + \mathbf{r}$ must be parallel to $\mathbf{p}$.
Therefore,
$3\mathbf{q} + \mathbf{r} = k\mathbf{p}$
for some scalar $k \in \mathbb{R}$.

1. Express the position vector $\mathbf{c}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ using the ratio theorem.

Since $B$ divides $AC$ in the ratio $1:3$, by the ratio theorem:
$\mathbf{b} = \frac{3\mathbf{a} + 1\mathbf{c}}{1 + 3}$
$4\mathbf{b} = 3\mathbf{a} + \mathbf{c}$
$\mathbf{c} = 4\mathbf{b} - 3\mathbf{a}$

2. Express the vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

Since $D$ lies on $OB$ produced and $OB : OD = 1 : m$, we have $\mathbf{d} = m\mathbf{b}$.

$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$

$\overrightarrow{CD} = \mathbf{d} - \mathbf{c}$
$\overrightarrow{CD} = m\mathbf{b} - (4\mathbf{b} - 3\mathbf{a})$
$\overrightarrow{CD} = (m - 4)\mathbf{b} + 3\mathbf{a}$

3. Apply the condition for perpendicular vectors and expand the dot product.

Since $\overrightarrow{AB}$ is perpendicular to $\overrightarrow{CD}$, their scalar (dot) product is zero:
$\overrightarrow{AB} \cdot \overrightarrow{CD} = 0$
$(\mathbf{b} - \mathbf{a}) \cdot ((m - 4)\mathbf{b} + 3\mathbf{a}) = 0$
Expanding the expression:
$(m - 4)(\mathbf{b} \cdot \mathbf{b}) + 3(\mathbf{b} \cdot \mathbf{a}) - (m - 4)(\mathbf{a} \cdot \mathbf{b}) - 3(\mathbf{a} \cdot \mathbf{a}) = 0$

4. Use the properties of unit vectors to simplify and solve for $m$.

Given that $\mathbf{a}$ and $\mathbf{b}$ are unit vectors, $|\mathbf{a}| = 1 \implies \mathbf{a} \cdot \mathbf{a} = 1$ and $|\mathbf{b}| = 1 \implies \mathbf{b} \cdot \mathbf{b} = 1$.
$(m - 4)(1) + 3(\mathbf{a} \cdot \mathbf{b}) - (m - 4)(\mathbf{a} \cdot \mathbf{b}) - 3(1) = 0$
$m - 7 + (7 - m)(\mathbf{a} \cdot \mathbf{b}) = 0$
Factorising the expression:
$(m - 7)(1 - \mathbf{a} \cdot \mathbf{b}) = 0$
Since $\mathbf{a}$ and $\mathbf{b}$ are non-parallel unit vectors, the angle $\theta$ between them is not $0$, so $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta = \cos\theta \neq 1$.
Therefore, $1 - \mathbf{a} \cdot \mathbf{b} \neq 0$, which means:
$m - 7 = 0$
$m = 7$