已知 $\overrightarrow{OA} = 3\mathbf{i} + \mathbf{j} + 3\mathbf{k}$ 且 $\overrightarrow{OB} = 5\mathbf{i} - 4\mathbf{j} + 3\mathbf{k}$。求直线 $OB$ 上一点 $R$ 的位置向量，使得 $AR$ 垂直于 $OB$。由此求点 $A$ 关于直线 $OB$ 的对称点 $A'$ 的位置向量。

1. Express the position vector of point $R$ in terms of a scalar parameter, using the collinearity of $O$, $R$, and $B$.

Since $R$ lies on the line passing through $O$ and $B$, $\overrightarrow{OR}$ is a scalar multiple of $\overrightarrow{OB}$.
$\overrightarrow{OR} = \lambda \begin{pmatrix} 5 \\ -4 \\ 3 \end{pmatrix} = \begin{pmatrix} 5\lambda \\ -4\lambda \\ 3\lambda \end{pmatrix}$

2. Find the value of the parameter $\lambda$ using the perpendicular condition between $AR$ and $OB$.

The direction vector $\overrightarrow{AR}$ is given by:
$\overrightarrow{AR} = \overrightarrow{OR} - \overrightarrow{OA} = \begin{pmatrix} 5\lambda - 3 \\ -4\lambda - 1 \\ 3\lambda - 3 \end{pmatrix}$
Since $AR \perp OB$, their scalar product must be equal to zero:
$\overrightarrow{AR} \cdot \overrightarrow{OB} = 0$
$\begin{pmatrix} 5\lambda - 3 \\ -4\lambda - 1 \\ 3\lambda - 3 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ -4 \\ 3 \end{pmatrix} = 0$
$5(5\lambda - 3) - 4(-4\lambda - 1) + 3(3\lambda - 3) = 0$
$25\lambda - 15 + 16\lambda + 4 + 9\lambda - 9 = 0$
$50\lambda - 20 = 0 \implies \lambda = \frac{2}{5}$

3. Substitute $\lambda$ to find the specific position vector of $R$.

$\overrightarrow{OR} = \frac{2}{5} \begin{pmatrix} 5 \\ -4 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ -8/5 \\ 6/5 \end{pmatrix}$
Thus, the position vector is $\overrightarrow{OR} = 2\mathbf{i} - \frac{8}{5}\mathbf{j} + \frac{6}{5}\mathbf{k}$.

4. Find the position vector of the reflection point $A'$ using the midpoint property.

The foot of the perpendicular $R$ is the midpoint of the line segment connecting $A$ and its reflection $A'$.
$\overrightarrow{OR} = \frac{\overrightarrow{OA} + \overrightarrow{OA'}}{2}$
$\overrightarrow{OA'} = 2\overrightarrow{OR} - \overrightarrow{OA}$
$\overrightarrow{OA'} = 2 \begin{pmatrix} 2 \\ -8/5 \\ 6/5 \end{pmatrix} - \begin{pmatrix} 3 \\ 1 \\ 3 \end{pmatrix}$
$\overrightarrow{OA'} = \begin{pmatrix} 4 \\ -16/5 \\ 12/5 \end{pmatrix} - \begin{pmatrix} 3 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ -21/5 \\ -3/5 \end{pmatrix}$
Thus, the position vector is $\overrightarrow{OA'} = \mathbf{i} - \frac{21}{5}\mathbf{j} - \frac{3}{5}\mathbf{k}$.