Referred to the origin O, let P, Q and R be distinct points with position vectors $\mathbf{p}$, $\mathbf{q}$ and $\mathbf{r}$ respectively.

1. Expand the given cross product expression using the distributive property.

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

2. Apply properties of the cross product to simplify the expression.

Using the properties $\mathbf{r}\times\mathbf{r} = \mathbf{0}$ and $\mathbf{a}\times\mathbf{b} = -\mathbf{b}\times\mathbf{a}$:

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

1. Rewrite the expression using the vector identity proven in part (a).

From part (a), we know that $\mathbf{p}\times\mathbf{q} + \mathbf{q}\times\mathbf{r} + \mathbf{r}\times\mathbf{p} = (\mathbf{r}-\mathbf{p})\times(\mathbf{r}-\mathbf{q})$.
Substituting this into the given expression:
$\frac{1}{2}|\mathbf{p}\times\mathbf{q}+\mathbf{q}\times\mathbf{r}+\mathbf{r}\times\mathbf{p}| = \frac{1}{2}|(\mathbf{r}-\mathbf{p})\times(\mathbf{r}-\mathbf{q})|$

2. Interpret the resulting vector product geometrically.

Identify the vectors inside the magnitude:
$\mathbf{r}-\mathbf{p} = \vec{PR}$
$\mathbf{r}-\mathbf{q} = \vec{QR}$

So the expression simplifies to $\frac{1}{2}|\vec{PR} \times \vec{QR}|$.
Since the magnitude of the cross product of two adjacent sides of a triangle gives the area of the parallelogram formed by them, half of this magnitude represents the area of the triangle itself.
Therefore, the expression gives the area of triangle $PQR$.

1. Deduce that points $P$, $Q$ and $R$ are collinear using the result from part (a).

From part (a), we know $(\mathbf{r}-\mathbf{p})\times(\mathbf{r}-\mathbf{q}) = \mathbf{p}\times\mathbf{q} + \mathbf{q}\times\mathbf{r} + \mathbf{r}\times\mathbf{p}$.
Given $\mathbf{p}\times\mathbf{q} + \mathbf{q}\times\mathbf{r} + \mathbf{r}\times\mathbf{p} = \mathbf{0}$, we have:
$\vec{PR} \times \vec{QR} = \mathbf{0}$
This implies that the vectors $\vec{PR}$ and $\vec{QR}$ are parallel. Since they share the common point $R$, points $P$, $Q$ and $R$ must be collinear.

2. Determine whether $R$ divides $PQ$ internally or externally.

Since $P$, $Q$ and $R$ are collinear, we consider their relative positions:

• If $P-Q-R$ (external division), then $PR = PQ + QR$. Given $PR = 3QR$, we get $3QR = PQ + QR \implies PQ = 2QR$. This means $PQ < PR$, which contradicts the given condition $PQ > PR$.
• If $R-P-Q$, then $QR = RP + PQ \implies QR = 3QR + PQ \implies PQ = -2QR$, which is impossible for positive lengths.
• If $P-R-Q$ (internal division), then $PQ = PR + RQ = 3QR + QR = 4QR$. Here $PQ > PR$ is satisfied since $4QR > 3QR$.
  Thus, $R$ must lie between $P$ and $Q$.

3. Use the ratio theorem to express $\mathbf{r}$.

Since $R$ lies between $P$ and $Q$ and $PR = 3QR$, $R$ divides the line segment $PQ$ internally in the ratio $3:1$.
Applying the ratio theorem:
$\mathbf{r} = \frac{1(\mathbf{p}) + 3(\mathbf{q})}{3 + 1} = \frac{1}{4}\mathbf{p} + \frac{3}{4}\mathbf{q}$