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 vector product on the left-hand side using the distributive law.

$\begin{aligned} \text{LHS} &= (\mathbf{r}-\mathbf{p})\times(\mathbf{r}-\mathbf{q}) \\ &= \mathbf{r}\times\mathbf{r} - \mathbf{r}\times\mathbf{q} - \mathbf{p}\times\mathbf{r} + \mathbf{p}\times\mathbf{q} \end{aligned}$

2. Apply the properties of vector products to simplify the expression to the right-hand side.

$\begin{aligned} &\text{Since } \mathbf{r}\times\mathbf{r} = \mathbf{0}, \; \mathbf{r}\times\mathbf{q} = -\mathbf{q}\times\mathbf{r} \text{ and } \mathbf{p}\times\mathbf{r} = -\mathbf{r}\times\mathbf{p}, \\ \text{LHS} &= \mathbf{0} - (-\mathbf{q}\times\mathbf{r}) - (-\mathbf{r}\times\mathbf{p}) + \mathbf{p}\times\mathbf{q} \\ &= \mathbf{q}\times\mathbf{r} + \mathbf{r}\times\mathbf{p} + \mathbf{p}\times\mathbf{q} \\ &= \mathbf{p}\times\mathbf{q} + \mathbf{q}\times\mathbf{r} + \mathbf{r}\times\mathbf{p} \\ &= \text{RHS (shown)} \end{aligned}$

1. Substitute the identity proved in part (a) into the given expression.

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, we get:
$\frac{1}{2}\,|\mathbf{p}\times\mathbf{q} + \mathbf{q}\times\mathbf{r} + \mathbf{r}\times\mathbf{p}| = \frac{1}{2}\,|(\mathbf{r}-\mathbf{p})\times(\mathbf{r}-\mathbf{q})|$

2. Interpret the vectors geometrically to find the meaning of the expression.

The vectors can be written in terms of the points $P$, $Q$, and $R$:
$\mathbf{r}-\mathbf{p} = \vec{PR}$
$\mathbf{r}-\mathbf{q} = \vec{QR}$

The expression becomes $\frac{1}{2}\,|\vec{PR}\times\vec{QR}|$.
The magnitude of the cross product of two vectors representing adjacent sides of a triangle gives the area of the parallelogram formed by them. Multiplying by $\frac{1}{2}$ gives the area of the triangle.

Therefore, the expression represents the area of triangle $PQR$.

1. Establish that points P, Q, and R are collinear.

From the result shown in (a), we substitute the given condition $\mathbf{p}\times\mathbf{q} + \mathbf{q}\times\mathbf{r} + \mathbf{r}\times\mathbf{p} = \mathbf{0}$ to obtain:
$(\mathbf{r}-\mathbf{p})\times(\mathbf{r}-\mathbf{q}) = \mathbf{0}$
$\vec{PR} \times \vec{QR} = \mathbf{0}$
Since the cross product of these two vectors is the zero vector, $\vec{PR}$ and $\vec{QR}$ are parallel. As they share the common point $R$, the points $P$, $Q$ and $R$ are collinear.

2. Set up the vector equations for the two possible cases and eliminate the invalid one.

Since the points are collinear and their lengths relate by $PR = 3QR$, the vectors must satisfy either $\vec{PR} = 3\vec{QR}$ or $\vec{PR} = -3\vec{QR}$.

Consider Case 1 where $\vec{PR} = 3\vec{QR}$:
$\mathbf{r} - \mathbf{p} = 3(\mathbf{r} - \mathbf{q})$
$2\mathbf{r} = 3\mathbf{q} - \mathbf{p} \implies \mathbf{r} = \frac{3}{2}\mathbf{q} - \frac{1}{2}\mathbf{p}$
We check the magnitude condition $PQ > PR$ for this case:
$\vec{PR} = \left(\frac{3}{2}\mathbf{q} - \frac{1}{2}\mathbf{p}\right) - \mathbf{p} = \frac{3}{2}(\mathbf{q} - \mathbf{p}) = \frac{3}{2}\vec{PQ}$
Taking magnitudes gives $PR = \frac{3}{2}PQ$, meaning $PQ < PR$. This contradicts the given condition $PQ > PR$, so we reject this case.

3. Solve the valid case to find $\mathbf{r}$ in terms of $\mathbf{p}$ and $\mathbf{q}$.

Consider Case 2 where $\vec{PR} = -3\vec{QR}$:
$\mathbf{r} - \mathbf{p} = -3(\mathbf{r} - \mathbf{q})$
$\mathbf{r} - \mathbf{p} = -3\mathbf{r} + 3\mathbf{q}$
$4\mathbf{r} = \mathbf{p} + 3\mathbf{q}$
$\mathbf{r} = \frac{1}{4}\mathbf{p} + \frac{3}{4}\mathbf{q}$

(Self-check for this case: $\vec{PR} = \mathbf{r} - \mathbf{p} = \frac{3}{4}\mathbf{q} - \frac{3}{4}\mathbf{p} = \frac{3}{4}\vec{PQ}$, so $PR = \frac{3}{4}PQ$, which correctly satisfies $PQ > PR$.)