在以原点 $O$ 为参考点的空间中，点 $A$ 和点 $B$ 满足 $\overrightarrow{OA} = \mathbf{a}$ 且 $\overrightarrow{OB} = \mathbf{b}$，其中 $\mathbf{a}$ 与 $\mathbf{b}$ 不平行。点 $P$ 在 $AB$ 的延长线上，满足 $AP : AB = 3 : 1$，且点 $Q$ 的位置向量为 $2\mathbf{a}$。

1. Find the position vector of $P$ using the given ratio.

Since $P$ is on $AB$ produced with $AP : AB = 3 : 1$, the points are collinear in the order $A-B-P$.
Thus, $B$ divides $AP$ in the ratio $1 : 2$.
Using the ratio theorem:
$\overrightarrow{OB} = \frac{2\overrightarrow{OA} + 1\overrightarrow{OP}}{1 + 2}$
$\mathbf{b} = \frac{2\mathbf{a} + \overrightarrow{OP}}{3}$
$3\mathbf{b} = 2\mathbf{a} + \overrightarrow{OP} \implies \overrightarrow{OP} = 3\mathbf{b} - 2\mathbf{a}$

2. Formulate the vector equations for lines $OB$ and $PQ$.

Line $OB$ passes through the origin and point $B$:
$l_{OB}: \mathbf{r} = \mu \mathbf{b}$
Line $PQ$ passes through $P$ and $Q(2\mathbf{a})$:
$l_{PQ}: \mathbf{r} = \overrightarrow{OP} + \lambda (\overrightarrow{OQ} - \overrightarrow{OP})$
$\mathbf{r} = (3\mathbf{b} - 2\mathbf{a}) + \lambda (2\mathbf{a} - (3\mathbf{b} - 2\mathbf{a}))$
$\mathbf{r} = (3\mathbf{b} - 2\mathbf{a}) + \lambda (4\mathbf{a} - 3\mathbf{b})$
$\mathbf{r} = (4\lambda - 2)\mathbf{a} + (3 - 3\lambda)\mathbf{b}$

3. Equate the line equations to find the intersection point parameters.

At the point of intersection, the position vectors from both line equations must be equal:
$\mu \mathbf{b} = (4\lambda - 2)\mathbf{a} + (3 - 3\lambda)\mathbf{b}$
Since $\mathbf{a}$ and $\mathbf{b}$ are non-parallel vectors, their coefficients can be equated independently:
For $\mathbf{a}$:
$4\lambda - 2 = 0 \implies \lambda = \frac{1}{2}$
For $\mathbf{b}$:
$\mu = 3 - 3\lambda$
$\mu = 3 - 3\left(\frac{1}{2}\right) = \frac{3}{2}$

4. State the final position vector of the intersection.

Substitute $\mu = \frac{3}{2}$ into $l_{OB}$ (or $\lambda = \frac{1}{2}$ into $l_{PQ}$) to get the position vector:
$\overrightarrow{OR} = \frac{3}{2}\mathbf{b}$

1. Identify the normal vector to the plane $OAB$ and establish the equation of the plane.

The normal vector to the plane $OAB$ is given by the cross product of two non-parallel vectors on the plane. Since $\mathbf{a}$ and $\mathbf{b}$ lie on the plane,

$\mathbf{n} = \mathbf{a} \times \mathbf{b} = 2\mathbf{i} - 4\mathbf{j} + 2\mathbf{k} = 2 \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$

We can use a simpler normal vector $\mathbf{n}' = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$.
Since the plane passes through the origin $O$, the scalar product equation of the plane is:
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = 0$

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

Let the foot of the perpendicular be $F$. The line $CF$ passes through $C(0, 3, 4)$ and is parallel to the normal vector of the plane.

Equation of line $CF$:
$\mathbf{r} = \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}, \quad \lambda \in \mathbb{R}$

3. Substitute the line equation into the plane equation to find the value of the parameter $\lambda$ at the intersection point.

At the point of intersection $F$, the line meets the plane. Substituting the line's vector equation into the plane's equation:

$\left[ \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \right] \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = 0$
$\begin{pmatrix} \lambda \\ 3 - 2\lambda \\ 4 + \lambda \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = 0$
$\lambda(1) + (3 - 2\lambda)(-2) + (4 + \lambda)(1) = 0$
$\lambda - 6 + 4\lambda + 4 + \lambda = 0$
$6\lambda - 2 = 0 \implies \lambda = \frac{1}{3}$

4. Substitute the value of $\lambda$ back into the line equation to find the position vector and state the coordinates of $F$.

Substitute $\lambda = \frac{1}{3}$ into the equation of line $CF$:

$\overrightarrow{OF} = \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix} + \frac{1}{3} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1/3 \\ 3 - 2/3 \\ 4 + 1/3 \end{pmatrix} = \begin{pmatrix} 1/3 \\ 7/3 \\ 13/3 \end{pmatrix}$

The coordinates of the foot of the perpendicular are $\left( \frac{1}{3}, \frac{7}{3}, \frac{13}{3} \right)$.