四面体 $OABC$ 具有四个三角形面，顶点 $O$ 位于原点。$A$、$B$ 和 $C$ 的坐标分别为 $(9, 0, 0)$、$(0, 2, 0)$ 和 $(0, 0, 3)$（见图）。

1. Find two non-parallel direction vectors lying on the plane.

$\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} - \begin{pmatrix} 9 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} -9 \\ 2 \\ 0 \end{pmatrix}$

$\vec{AC} = \vec{OC} - \vec{OA} = \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} - \begin{pmatrix} 9 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} -9 \\ 0 \\ 3 \end{pmatrix}$

2. Calculate the cross product to find a normal vector to the plane.

$\mathbf{n} = \vec{AB} \times \vec{AC} = \begin{pmatrix} -9 \\ 2 \\ 0 \end{pmatrix} \times \begin{pmatrix} -9 \\ 0 \\ 3 \end{pmatrix} = \begin{pmatrix} (2)(3) - (0)(0) \\ (0)(-9) - (-9)(3) \\ (-9)(0) - (2)(-9) \end{pmatrix} = \begin{pmatrix} 6 \\ 27 \\ 18 \end{pmatrix}$

We can take a simpler normal vector by dividing by $3$:
$\mathbf{n}_1 = \begin{pmatrix} 2 \\ 9 \\ 6 \end{pmatrix}$

3. Form the cartesian equation of the plane using the normal vector and a point on the plane.

Using point $A(9, 0, 0)$ and the normal vector $\mathbf{n}_1$:
$\mathbf{r} \cdot \mathbf{n}_1 = \vec{OA} \cdot \mathbf{n}_1$
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 9 \\ 6 \end{pmatrix} = \begin{pmatrix} 9 \\ 0 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 9 \\ 6 \end{pmatrix}$
$2x + 9y + 6z = 18$

1. Find the normal vector to face $ABC$ by deriving its Cartesian equation from its axial intercepts (as required from part a).

The plane intercepts the coordinate axes at $x=9$, $y=2$, and $z=3$.
The Cartesian equation using the intercept form is:
$\frac{x}{9} + \frac{y}{2} + \frac{z}{3} = 1$
Multiplying by $18$ gives:
$2x + 9y + 6z = 18$
Thus, the normal vector to the plane is $\mathbf{n} = \begin{pmatrix} 2 \\ 9 \\ 6 \end{pmatrix}$.

2. Calculate the acute angle between the $z$-axis and face $ABC$.

The direction vector of the $z$-axis is $\mathbf{k} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.
Let the acute angle between the $z$-axis and the plane be $\theta$. Using the relation between the plane's normal and the line's direction vector:
$\sin \theta = \frac{|\mathbf{n} \cdot \mathbf{k}|}{|\mathbf{n}|\,|\mathbf{k}|}$
$\sin \theta = \frac{\left| \begin{pmatrix} 2 \\ 9 \\ 6 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right|}{\sqrt{2^2 + 9^2 + 6^2} \sqrt{0^2 + 0^2 + 1^2}}$
$\sin \theta = \frac{6}{\sqrt{121} (1)} = \frac{6}{11}$
$\theta = \sin^{-1}\left(\frac{6}{11}\right) \approx 33.1^\circ \text{ (to 1 d.p.)}$

1. Formulate the equation of the perpendicular line from $O$ to the plane $ABC$.

From part (a), using the intercepts, the cartesian equation of the plane $ABC$ is $\frac{x}{9} + \frac{y}{2} + \frac{z}{3} = 1 \implies 2x + 9y + 6z = 18$.
The normal vector to the plane is $\mathbf{n} = \begin{pmatrix} 2 \\ 9 \\ 6 \end{pmatrix}$.
The line passing through the origin $O$ perpendicular to the plane has equation $\mathbf{r} = \lambda \begin{pmatrix} 2 \\ 9 \\ 6 \end{pmatrix}$.

2. Substitute the line equation into the plane equation to find the foot of the perpendicular.

Let the foot of the perpendicular be $F$. Then $\vec{OF} = \lambda \begin{pmatrix} 2 \\ 9 \\ 6 \end{pmatrix}$.
Substituting this into the plane's equation to find the intersection point:
$2(2\lambda) + 9(9\lambda) + 6(6\lambda) = 18$
$\lambda(4 + 81 + 36) = 18$
$121\lambda = 18 \implies \lambda = \frac{18}{121}$
Hence, $\vec{OF} = \frac{18}{121} \begin{pmatrix} 2 \\ 9 \\ 6 \end{pmatrix} = \begin{pmatrix} \frac{36}{121} \\ \frac{162}{121} \\ \frac{108}{121} \end{pmatrix}$.

1. Find the position vector of $O'$, the reflection of $O$ in face $ABC$.

From part (c), the coordinates of $F$, the foot of the perpendicular from $O$ to face $ABC$, are $\left(\frac{36}{121}, \frac{162}{121}, \frac{108}{121}\right)$.
Since $F$ is the midpoint of $O$ and its reflection $O'$, we have:
$\vec{OO'} = 2\vec{OF} = 2 \begin{pmatrix} 36/121 \\ 162/121 \\ 108/121 \end{pmatrix} = \begin{pmatrix} 72/121 \\ 324/121 \\ 216/121 \end{pmatrix}$

2. Find the midpoint $M$ of $AC$ and recognize its role in the reflection.

The line $s$ passes through $O$ and the midpoint of $AC$. Let this midpoint be $M$.
$\vec{OM} = \frac{1}{2}(\vec{OA} + \vec{OC}) = \frac{1}{2} \left[ \begin{pmatrix} 9 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} \right] = \begin{pmatrix} 9/2 \\ 0 \\ 3/2 \end{pmatrix}$
Since $M$ lies on the line segment $AC$, it is on the plane $ABC$. Therefore, $M$ is its own reflection in the plane. The reflected line must therefore pass through $O'$ and $M$.

3. Determine the direction vector of the reflected line and find the value of $a$.

The direction vector of the reflected line is:
$\vec{O'M} = \vec{OM} - \vec{OO'} = \begin{pmatrix} 9/2 \\ 0 \\ 3/2 \end{pmatrix} - \begin{pmatrix} 72/121 \\ 324/121 \\ 216/121 \end{pmatrix} = \begin{pmatrix} 945/242 \\ -648/242 \\ -69/242 \end{pmatrix}$
Factoring out the scalar $\frac{3}{242}$ to match the integer components given in the question:
$\vec{O'M} = \frac{3}{242} \begin{pmatrix} 315 \\ -216 \\ -23 \end{pmatrix}$
Using $\vec{OM}$ as the position vector and $\begin{pmatrix} 315 \\ -216 \\ -23 \end{pmatrix}$ as the direction vector, the equation of the reflected line is:
$\mathbf{r} = \begin{pmatrix} 9/2 \\ 0 \\ 3/2 \end{pmatrix} + \mu \begin{pmatrix} 315 \\ -216 \\ -23 \end{pmatrix}$
Comparing this with the given equation $\mathbf{r} = \begin{pmatrix} 9/2 \\ 0 \\ 3/2 \end{pmatrix} + \mu \begin{pmatrix} a \\ -216 \\ -23 \end{pmatrix}$, we obtain $a = 315$.