相对于原点 $O$，点 $A$ 和点 $B$ 分别具有非零且不平行的位置向量 $\mathbf{a}$ 和 $\mathbf{b}$。平面 $p$ 的方程为 $\mathbf{r} \cdot \mathbf{n} = 0$。

1. Express the dot product of $\overrightarrow{AB}$ and $\mathbf{n}$ in terms of given position vectors to show it equals 0.

Using the position vectors, $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$.
Taking the scalar product with $\mathbf{n}$:
$\overrightarrow{AB} \cdot \mathbf{n} = (\mathbf{b} - \mathbf{a}) \cdot \mathbf{n}$
$\overrightarrow{AB} \cdot \mathbf{n} = \mathbf{b} \cdot \mathbf{n} - \mathbf{a} \cdot \mathbf{n}$
Since we are given that $\mathbf{a} \cdot \mathbf{n} = \mathbf{b} \cdot \mathbf{n}$, it follows that:
$\overrightarrow{AB} \cdot \mathbf{n} = 0$
Since their scalar product is zero, $\overrightarrow{AB}$ is perpendicular to $\mathbf{n}$.

2. Deduce the geometrical relationship between $\overrightarrow{AB}$ and the plane $p$.

The vector $\mathbf{n}$ is the normal vector to the plane $p$.
Since $\overrightarrow{AB}$ is perpendicular to the normal of the plane, $\overrightarrow{AB}$ must be parallel to the plane $p$.

1. Identify a point on the plane $p$ and write the equation of the line

The equation of the plane $p$ is $\mathbf{r} \cdot \mathbf{n} = 0$. Since $\mathbf{0} \cdot \mathbf{n} = 0$, the origin $O$ lies on the plane $p$.
Since the line must be parallel to $\overrightarrow{AB}$ (which is $\mathbf{b} - \mathbf{a}$) and contained in $p$, we can construct the line passing through $O$ with direction $\mathbf{b} - \mathbf{a}$.

The equation of the line is:
$\mathbf{r} = \lambda (\mathbf{b} - \mathbf{a}), \quad \lambda \in \mathbb{R}$

1. Formulate the vector equation of the line passing through $C$ and perpendicular to the plane.

Let the line passing through $C$ and perpendicular to plane $p$ be $l$. Since $l$ is perpendicular to the plane, its direction vector is the normal vector $\mathbf{n}$.
$l: \mathbf{r} = \mathbf{c} + \lambda \mathbf{n}, \text{ for } \lambda \in \mathbb{R}$

2. Find the parameter $\lambda$ at the intersection of the line and the plane $p$.

Since $F$ is the foot of the perpendicular, it lies on both the line $l$ and the plane $p$. Substituting the line equation into the plane equation $\mathbf{r} \cdot \mathbf{n} = 0$:
$(\mathbf{c} + \lambda \mathbf{n}) \cdot \mathbf{n} = 0$
$\mathbf{c} \cdot \mathbf{n} + \lambda (\mathbf{n} \cdot \mathbf{n}) = 0$
Since $\mathbf{n} \cdot \mathbf{n} = |\mathbf{n}|^2$:
$\lambda = -\frac{\mathbf{c} \cdot \mathbf{n}}{|\mathbf{n}|^2}$

3. Determine the position vector of $F$.

Substitute the expression for $\lambda$ back into the equation of the line to find the position vector of $F$:
$\overrightarrow{OF} = \mathbf{c} - \frac{\mathbf{c} \cdot \mathbf{n}}{|\mathbf{n}|^2} \mathbf{n}$