平面 $P_1, P_2$ 的方程为 $P_1:\quad \mathbf{r} = \begin{pmatrix}-3 \\ 10 \\ 3\end{pmatrix} + \lambda\begin{pmatrix}-2 \\ 12 \\ 3\end{pmatrix} + \beta\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix},\quad \lambda,\beta \in \mathbb{R}$ $P_2:\quad \mathbf{r} \cdot \begin{pmatrix}-1 \\ 0 \\ 1\end{pmatrix} = 1$

1. Formulate the vector equation of the line passing through $A$ and perpendicular to $P_2$.

The normal vector of $P_2$ is $\mathbf{n} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$.
Let $l$ be the line passing through $A$ and perpendicular to $P_2$. Its vector equation is
$\mathbf{r} = \vec{OA} + \mu \mathbf{n} = \begin{pmatrix} -3 \\ 10 \\ 3 \end{pmatrix} + \mu \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}, \quad \mu \in \mathbb{R}$

2. Substitute the line equation into $P_2$ to find the value of $\mu$ at the intersection point.

At the point of intersection $F$, the line satisfies the equation of $P_2$:
$\left[ \begin{pmatrix} -3 \\ 10 \\ 3 \end{pmatrix} + \mu \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \right] \cdot \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} = 1$
$\begin{pmatrix} -3-\mu \\ 10 \\ 3+\mu \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} = 1$
$(-3-\mu)(-1) + 10(0) + (3+\mu)(1) = 1$
$3 + \mu + 3 + \mu = 1$
$6 + 2\mu = 1$
$2\mu = -5 \implies \mu = -2.5$

3. Evaluate the position vector of the foot of the perpendicular, $F$.

Substitute $\mu = -2.5$ back into the line equation:
$\vec{OF} = \begin{pmatrix} -3 \\ 10 \\ 3 \end{pmatrix} - 2.5 \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -0.5 \\ 10 \\ 0.5 \end{pmatrix}$
The coordinates of the foot of the perpendicular are $(-0.5, 10, 0.5)$.

4. Show that $B$ is the reflection of $A$ in $P_2$.

Let $B'$ be the reflection of $A$ in $P_2$. Since $F$ must be the midpoint of $A$ and its reflection, we have $\vec{OB'} = \vec{OF} + \vec{AF}$.
First, compute the vector from $A$ to $F$:
$\vec{AF} = \vec{OF} - \vec{OA} = \begin{pmatrix} -0.5 \\ 10 \\ 0.5 \end{pmatrix} - \begin{pmatrix} -3 \\ 10 \\ 3 \end{pmatrix} = \begin{pmatrix} 2.5 \\ 0 \\ -2.5 \end{pmatrix}$
Now find the position vector of $B'$:
$\vec{OB'} = \vec{OF} + \vec{AF} = \begin{pmatrix} -0.5 \\ 10 \\ 0.5 \end{pmatrix} + \begin{pmatrix} 2.5 \\ 0 \\ -2.5 \end{pmatrix} = \begin{pmatrix} 2 \\ 10 \\ -2 \end{pmatrix}$
Since $\vec{OB'}$ matches the position vector of $B(2, 10, -2)$ exactly, point $B$ is indeed the reflection of $A$ in $P_2$.

1. Substitute the parametric vector equation of plane $P_1$ into the scalar product equation of plane $P_2$.

$\left( \begin{pmatrix} -3 \\ 10 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} -2 \\ 12 \\ 3 \end{pmatrix} + \beta\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right) \cdot \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} = 1$

2. Expand the scalar product to find a relationship between the parameters $\lambda$ and $\beta$.

$\begin{pmatrix} -3 \\ 10 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} -2 \\ 12 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} + \beta \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} = 1$
$(3 + 0 + 3) + \lambda(2 + 0 + 3) + \beta(-1 + 0 + 1) = 1$
$6 + 5\lambda + 0 = 1$
$5\lambda = -5 \implies \lambda = -1$

3. Substitute $\lambda = -1$ back into the equation of $P_1$ to determine the vector equation of the line of intersection.

$\mathbf{r} = \begin{pmatrix} -3 \\ 10 \\ 3 \end{pmatrix} - \begin{pmatrix} -2 \\ 12 \\ 3 \end{pmatrix} + \beta\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$
$\mathbf{r} = \begin{pmatrix} -1 \\ -2 \\ 0 \end{pmatrix} + \beta\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad \beta \in \mathbb{R}$

1. Compute the vector equation of the intersection line $L$ (required from Part 1).

Substitute the parametric equation of $P_1$ into the cartesian equation of $P_2$ to find their intersection:
$\left[ \begin{pmatrix} -3 \\ 10 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} -2 \\ 12 \\ 3 \end{pmatrix} + \beta \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right] \cdot \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} = 1$
$(3 + 3) + \lambda(2 + 3) + \beta(-1 + 1) = 1$
$6 + 5\lambda = 1 \implies \lambda = -1$
Substitute $\lambda = -1$ into $P_1$ to obtain the equation of line $L$:
$\mathbf{r} = \begin{pmatrix} -3 \\ 10 \\ 3 \end{pmatrix} - \begin{pmatrix} -2 \\ 12 \\ 3 \end{pmatrix} + \beta \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ -2 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$

2. Deduce points and directions on the reflected plane $P_3$.

The point $A(-3, 10, 3)$ lies on $P_1$ (it is the position vector when $\lambda=\beta=0$). From Part 0, its reflection in $P_2$ is $B(2, 10, -2)$. Because $P_3$ is the reflection of $P_1$ in $P_2$, $B$ must lie on $P_3$.

The line $L$ lies on both $P_1$ and $P_2$, so it is invariant under reflection in $P_2$ and must also lie completely on $P_3$.

A second direction vector for $P_3$ is the vector connecting the base point on $L$, $C(-1, -2, 0)$, to the point $B(2, 10, -2)$:
$\vec{CB} = \begin{pmatrix} 2 \\ 10 \\ -2 \end{pmatrix} - \begin{pmatrix} -1 \\ -2 \\ 0 \end{pmatrix} = \begin{pmatrix} 3 \\ 12 \\ -2 \end{pmatrix}$

3. Find a vector perpendicular to $P_3$.

$P_3$ is parallel to $L$'s direction $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ and to $\vec{CB}$. We find the normal vector by taking their cross product:
$\mathbf{n}_3 = \begin{pmatrix} 3 \\ 12 \\ -2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} (12)(1) - (-2)(1) \\ (-2)(1) - (3)(1) \\ (3)(1) - (12)(1) \end{pmatrix} = \begin{pmatrix} 14 \\ -5 \\ -9 \end{pmatrix}$

4. Formulate the scalar product equation of $P_3$.

Using the point $C(-1, -2, 0)$ on $P_3$ and the normal vector $\mathbf{n}_3$:
$\mathbf{r} \cdot \begin{pmatrix} 14 \\ -5 \\ -9 \end{pmatrix} = \begin{pmatrix} -1 \\ -2 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 14 \\ -5 \\ -9 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 14 \\ -5 \\ -9 \end{pmatrix} = -14 + 10 + 0 = -4$