相对于原点 $O$，点 $A$、$B$ 和 $C$ 的位置向量分别为 $6\mathbf{i}$、$4\mathbf{i}+2\mathbf{j}+4\mathbf{k}$ 和 $3\mathbf{i}+2\mathbf{j}+6\mathbf{k}$。平面 $p_{1}$ 的方程为 $x - y + z = 1$。

1. Find two direction vectors parallel to the plane.

$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 4 \\ 2 \\ 4 \end{pmatrix} - \begin{pmatrix} 6 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ 2 \\ 4 \end{pmatrix}$

$\overrightarrow{AC} = \mathbf{c} - \mathbf{a} = \begin{pmatrix} 3 \\ 2 \\ 6 \end{pmatrix} - \begin{pmatrix} 6 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \\ 6 \end{pmatrix}$

2. Find a normal vector to the plane using the cross product.

$\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix} -2 \\ 2 \\ 4 \end{pmatrix} \times \begin{pmatrix} -3 \\ 2 \\ 6 \end{pmatrix} = \begin{pmatrix} (2)(6) - (4)(2) \\ (4)(-3) - (-2)(6) \\ (-2)(2) - (2)(-3) \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \\ 2 \end{pmatrix}$
Taking a simpler, parallel normal vector by dividing by 2, let $\mathbf{n}' = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$.

3. Formulate the equation of the plane in scalar product form.

The equation of the plane is given by $\mathbf{r} \cdot \mathbf{n}' = \mathbf{a} \cdot \mathbf{n}'$.
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 0 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = 12$

1. Find the equation of plane $ABC$ (required from part (i)).

$\vec{AB} = \begin{pmatrix} 4 \\ 2 \\ 4 \end{pmatrix} - \begin{pmatrix} 6 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ 2 \\ 4 \end{pmatrix}$

$\vec{AC} = \begin{pmatrix} 3 \\ 2 \\ 6 \end{pmatrix} - \begin{pmatrix} 6 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \\ 6 \end{pmatrix}$

Normal vector $\mathbf{n}_{ABC} = \vec{AB} \times \vec{AC} = \begin{pmatrix} -2 \\ 2 \\ 4 \end{pmatrix} \times \begin{pmatrix} -3 \\ 2 \\ 6 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \\ 2 \end{pmatrix}$

Using $\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$ as a simplified normal vector, the scalar product equation of plane $ABC$ is:

$\mathbf{r} \cdot \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 0 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = 12$

2. Formulate the vector equation of the line passing through $D$ and perpendicular to plane $p_1$.

The normal vector of plane $p_1$ ($x - y + z = 1$) is $\mathbf{n}_1 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$.

The equation of the line passing through $D(3,-1,4)$ and perpendicular to $p_1$ has this normal vector as its direction vector:
$l_{D}: \mathbf{r} = \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}, \text{ where } \lambda \in \mathbb{R}$

3. Substitute the line equation into the equation of plane $ABC$ to find the coordinates of intersection point $S$.

At the point of intersection $S$, the line $l_{D}$ meets the plane $ABC$:
$\left[ \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \right] \cdot \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = 12$

$\left( 3(2) + (-1)(0) + 4(1) \right) + \lambda \left( 1(2) + (-1)(0) + 1(1) \right) = 12$

$10 + 3\lambda = 12 \implies 3\lambda = 2 \implies \lambda = \frac{2}{3}$

Substitute $\lambda = \frac{2}{3}$ into the line equation:
$\vec{OS} = \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} + \frac{2}{3} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 11/3 \\ -5/3 \\ 14/3 \end{pmatrix}$

The coordinates of $S$ are $\left( \frac{11}{3}, -\frac{5}{3}, \frac{14}{3} \right)$.

1. Calculate the coordinates of $S$ by solving earlier parts (i) and (ii)

From (i), the normal to plane $ABC$ is $\vec{AB} \times \vec{AC} = \begin{pmatrix}-2\\2\\4\end{pmatrix} \times \begin{pmatrix}-3\\2\\6\end{pmatrix} = \begin{pmatrix}4\\0\\2\end{pmatrix} = 2\begin{pmatrix}2\\0\\1\end{pmatrix}$.
Equation of $ABC$: $\mathbf{r} \cdot \begin{pmatrix}2\\0\\1\end{pmatrix} = \begin{pmatrix}6\\0\\0\end{pmatrix} \cdot \begin{pmatrix}2\\0\\1\end{pmatrix} = 12$.
From (ii), the line from $D(3,-1,4)$ perpendicular to $p_1 (x-y+z=1)$ has the equation $\mathbf{r} = \begin{pmatrix}3\\-1\\4\end{pmatrix} + \lambda\begin{pmatrix}1\\-1\\1\end{pmatrix}$.
Substituting this line into the plane $ABC$:
$2(3+\lambda) + (4+\lambda) = 12 \implies 3\lambda = 2 \implies \lambda = \frac{2}{3}$.
Thus, $\vec{OS} = \begin{pmatrix}11/3\\-5/3\\14/3\end{pmatrix}$.

2. Identify the normal vector to the plane $p_2$

The plane $p_2$ is equidistant from $S$ and $D$, meaning it is the perpendicular bisector plane of the line segment $SD$.
Its normal vector $\mathbf{n}_2$ is parallel to $\vec{DS}$.
Since $S$ lies on the perpendicular dropped from $D$ to $p_1$, $\vec{DS}$ must be parallel to the normal of $p_1$.
Therefore, we can simply take $\mathbf{n}_2 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$.

3. Find the position vector of the midpoint of $SD$

Let $M$ be the midpoint of $SD$.
$\vec{OM} = \frac{\vec{OS} + \vec{OD}}{2}$
$\vec{OM} = \frac{1}{2} \left[ \begin{pmatrix} 11/3 \\ -5/3 \\ 14/3 \end{pmatrix} + \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} \right] = \frac{1}{2} \begin{pmatrix} 20/3 \\ -8/3 \\ 26/3 \end{pmatrix} = \begin{pmatrix} 10/3 \\ -4/3 \\ 13/3 \end{pmatrix}$

4. Form the equation of the plane $p_2$

The equation of $p_2$ is $\mathbf{r} \cdot \mathbf{n}_2 = \vec{OM} \cdot \mathbf{n}_2$.
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 10/3 \\ -4/3 \\ 13/3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = \frac{10}{3} - \left(-\frac{4}{3}\right) + \frac{13}{3} = \frac{27}{3} = 9$
The equation of the plane is $\mathbf{r} \cdot \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = 9$ (or $x - y + z = 9$).