点 P、Q 和 R 相对于原点 O 的位置向量分别为 p、q 和 r。向量 p 是单位向量，而向量 q 和 r 是非零向量。

1. State the geometrical meaning based on the property of the unit vector

Since $\mathbf{p}$ is a unit vector ($|\mathbf{p}| = 1$), the absolute value of the scalar product $|\mathbf{p}\cdot\mathbf{q}|$ gives the magnitude of the projection of $\mathbf{q}$ onto the direction of $\mathbf{p}$.

1. Express $\overrightarrow{OM}$ as a scalar multiple of $\mathbf{p}$ and set up the perpendicularity condition for $\overrightarrow{QM}$.

Since $M$ lies on the line $OP$, let $\overrightarrow{OM} = \lambda \mathbf{p}$ for some real number $\lambda$.

Since $M$ is the foot of the perpendicular from $Q$ to $OP$, the vector $\overrightarrow{QM}$ is perpendicular to $\mathbf{p}$.
Thus, $\overrightarrow{QM} \cdot \mathbf{p} = 0$.

2. Substitute $\overrightarrow{QM} = \overrightarrow{OM} - \overrightarrow{OQ}$, expand the dot product, and solve for $\lambda$.

$(\overrightarrow{OM} - \overrightarrow{OQ}) \cdot \mathbf{p} = 0$
$(\lambda \mathbf{p} - \mathbf{q}) \cdot \mathbf{p} = 0$
$\lambda (\mathbf{p} \cdot \mathbf{p}) - (\mathbf{q} \cdot \mathbf{p}) = 0$

Since $\mathbf{p}$ is a unit vector, $\mathbf{p} \cdot \mathbf{p} = |\mathbf{p}|^2 = 1$.
Given that $\mathbf{p} \cdot \mathbf{q} = 2$, we have:
$\lambda (1) - 2 = 0 \implies \lambda = 2$

Therefore, $\overrightarrow{OM} = 2\mathbf{p}$ (shown).

1. Express the vectors $\vec{MQ}$ and $\vec{MR}$ in terms of $\mathbf{p}$ and $\mathbf{q}$.

Using the position vectors given and the result $\vec{OM} = 2\mathbf{p}$:
$\vec{MQ} = \vec{OQ} - \vec{OM} = \mathbf{q} - 2\mathbf{p}$
$\vec{MR} = \vec{OR} - \vec{OM} = k(\mathbf{p} - \mathbf{q}) - 2\mathbf{p} = (k - 2)\mathbf{p} - k\mathbf{q}$

2. Set up the cross product $\vec{MQ} \times \vec{MR}$ for the area of the triangle and expand it.

The area of $\triangle RQM$ is $\frac{1}{2} |\vec{MQ} \times \vec{MR}|$.

Expanding the cross product:
$\vec{MQ} \times \vec{MR} = (\mathbf{q} - 2\mathbf{p}) \times ((k - 2)\mathbf{p} - k\mathbf{q})$
$= \mathbf{q} \times ((k - 2)\mathbf{p}) + \mathbf{q} \times (-k\mathbf{q}) - 2\mathbf{p} \times ((k - 2)\mathbf{p}) - 2\mathbf{p} \times (-k\mathbf{q})$

3. Simplify the cross product using its properties.

Since the cross product of a vector with itself is the zero vector ($\mathbf{p} \times \mathbf{p} = \mathbf{0}$ and $\mathbf{q} \times \mathbf{q} = \mathbf{0}$):
$\vec{MQ} \times \vec{MR} = (k - 2)(\mathbf{q} \times \mathbf{p}) + 2k(\mathbf{p} \times \mathbf{q})$

Using the anti-commutative property $\mathbf{q} \times \mathbf{p} = -(\mathbf{p} \times \mathbf{q})$:
$\vec{MQ} \times \vec{MR} = -(k - 2)(\mathbf{p} \times \mathbf{q}) + 2k(\mathbf{p} \times \mathbf{q})$
$= (-k + 2 + 2k)(\mathbf{p} \times \mathbf{q})$
$= (k + 2)(\mathbf{p} \times \mathbf{q})$

4. Determine the constant $a$.

$\text{Area} = \frac{1}{2} |(k + 2)(\mathbf{p} \times \mathbf{q})|$
$= \frac{1}{2} |k + 2| \, |\mathbf{p} \times \mathbf{q}|$

Comparing this with the given area $a\,|\mathbf{p} \times \mathbf{q}|$, we find:
$a = \frac{1}{2}|k + 2|$

5. Substitute $k = -2$ to describe the relationship between the points.

When $k = -2$, the constant $a = \frac{1}{2}|-2 + 2| = 0$.

Since the area of $\triangle RQM$ is $0$, the points $R$, $Q$ and $M$ are collinear.