以原点 $O$ 为基准，点 $A$ 和 $B$ 满足 $\overrightarrow{OA} = a$ 且 $\overrightarrow{OB} = b$，其中 $a$ 不平行于 $b$。点 $P$ 在 $AB$ 的延长线上，且满足 $AP:AB = 3:1$，点 $Q$ 的位置向量为 $2a$。

1. Find the position vector of $P$.

Since $P$ is on $AB$ produced with $AP:AB = 3:1$, we know that $\overrightarrow{AP} = 3\overrightarrow{AB}$.
$\overrightarrow{OP} - \overrightarrow{OA} = 3(\overrightarrow{OB} - \overrightarrow{OA})$
$\overrightarrow{OP} - \mathbf{a} = 3(\mathbf{b} - \mathbf{a})$
$\overrightarrow{OP} = 3\mathbf{b} - 2\mathbf{a}$

2. Express the position vector of a general point on the line $PQ$.

The line $PQ$ passes through $Q(2\mathbf{a})$ and $P(3\mathbf{b} - 2\mathbf{a})$. Let $R$ be the point of intersection. Since $R$ lies on $PQ$, its position vector can be written using a scalar parameter $\mu$:
$\overrightarrow{OR} = (1 - \mu)\overrightarrow{OQ} + \mu\overrightarrow{OP}$
$\overrightarrow{OR} = (1 - \mu)(2\mathbf{a}) + \mu(3\mathbf{b} - 2\mathbf{a})$
$\overrightarrow{OR} = (2 - 2\mu - 2\mu)\mathbf{a} + 3\mu\mathbf{b}$
$\overrightarrow{OR} = (2 - 4\mu)\mathbf{a} + 3\mu\mathbf{b}$

3. Set up an equation using the condition that the intersection point lies on $OB$.

Since the point $R$ also lies on the line $OB$, its position vector must be a scalar multiple of $\mathbf{b}$. Let $\overrightarrow{OR} = \lambda\mathbf{b}$ for some scalar $\lambda$.
Equating the two expressions for $\overrightarrow{OR}$:
$\lambda\mathbf{b} = (2 - 4\mu)\mathbf{a} + 3\mu\mathbf{b}$

4. Solve for the parameters to find the point of intersection.

Since vectors $\mathbf{a}$ and $\mathbf{b}$ are non-zero and not parallel, their coefficients on both sides of the equation must be equal.
Comparing the coefficients of $\mathbf{a}$:
$0 = 2 - 4\mu \implies \mu = \frac{1}{2}$
Comparing the coefficients of $\mathbf{b}$:
$\lambda = 3\mu \implies \lambda = 3\left(\frac{1}{2}\right) = \frac{3}{2}$
Therefore, the position vector of the point of intersection is $\frac{3}{2}\mathbf{b}$.

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

The normal to the plane $OAB$, $\mathbf{n}$, is parallel to the cross product of two non-parallel vectors on the plane.
$\mathbf{n} = a \times b = \begin{pmatrix} 2 \\ -4 \\ 2 \end{pmatrix}$
Since the plane contains the origin $O(0,0,0)$, the scalar product equation of the plane $OAB$ is:
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ -4 \\ 2 \end{pmatrix} = 0$

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

The perpendicular line $l$ passes through $C(0,3,4)$ and has a direction vector parallel to the normal $\mathbf{n}$.
$l: \mathbf{r} = \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -4 \\ 2 \end{pmatrix}, \text{ where } \lambda \in \mathbb{R}$

3. Find the value of the parameter $\lambda$ at the point of intersection between the line and the plane.

At the foot of the perpendicular $F$, the line intersects the plane. We substitute the line equation into the plane equation:
$\left[ \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -4 \\ 2 \end{pmatrix} \right] \cdot \begin{pmatrix} 2 \\ -4 \\ 2 \end{pmatrix} = 0$
$\begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -4 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -4 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -4 \\ 2 \end{pmatrix} = 0$
$(0 - 12 + 8) + \lambda(2^2 + (-4)^2 + 2^2) = 0$
$-4 + \lambda(4 + 16 + 4) = 0$
$-4 + 24\lambda = 0 \implies \lambda = \frac{1}{6}$

4. Calculate the coordinates of the foot of the perpendicular $F$.

Substitute $\lambda = \frac{1}{6}$ back into the equation of line $l$ to find the position vector of $F$:
$\overrightarrow{OF} = \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix} + \frac{1}{6} \begin{pmatrix} 2 \\ -4 \\ 2 \end{pmatrix}$
$\overrightarrow{OF} = \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix} + \begin{pmatrix} \frac{1}{3} \\ -\frac{2}{3} \\ \frac{1}{3} \end{pmatrix} = \begin{pmatrix} \frac{1}{3} \\ \frac{7}{3} \\ \frac{13}{3} \end{pmatrix}$

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