两个平面 $p_1$ 和 $p_2$ 的方程分别为 $$\begin{aligned} 2x + sy + z &= -3, \\ x - 2y + 3z &= 1, \end{aligned}$$ 其中 $s$ 是一个负实常数。

1. Identify the normal vectors of the two planes.

From the Cartesian equations of the planes, the normal vectors to $p_1$ and $p_2$ are respectively:
$\mathbf{n}_1 = \begin{pmatrix} 2 \\ s \\ 1 \end{pmatrix} \quad \text{and} \quad \mathbf{n}_2 = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}$

2. Calculate the cross product of the normal vectors.

A vector parallel to both planes must be perpendicular to both normal vectors. We can find this by taking their cross product:
$\mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} 2 \\ s \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}$
$= \begin{pmatrix} (s)(3) - (1)(-2) \\ (1)(1) - (2)(3) \\ (2)(-2) - (s)(1) \end{pmatrix}$
$= \begin{pmatrix} 3s + 2 \\ 1 - 6 \\ -4 - s \end{pmatrix} = \begin{pmatrix} 3s + 2 \\ -5 \\ -s - 4 \end{pmatrix}$

1. State the condition for a point to lie on the $xz$-plane.

For a point to lie on the $xz$-plane, its $y$-coordinate must be zero. Therefore, we set $y = 0$.

2. Substitute $y = 0$ into the equations of $p_1$ and $p_2$ to form a system of linear equations.

Substitute $y = 0$ into the equation for $p_1$:
$2x + s(0) + z = -3 \implies 2x + z = -3 \quad \text{--- (1)}$

Substitute $y = 0$ into the equation for $p_2$:
$x - 2(0) + 3z = 1 \implies x + 3z = 1 \quad \text{--- (2)}$

3. Solve the simultaneous equations for $x$ and $z$.

From (2), we can express $x$ in terms of $z$:
$x = 1 - 3z$

Substitute this into (1):
$2(1 - 3z) + z = -3$
$2 - 6z + z = -3$
$-5z = -5 \implies z = 1$

Substitute $z = 1$ back into the expression for $x$:
$x = 1 - 3(1) = -2$

4. State the final coordinates of point $A$.

Since $x = -2$, $y = 0$, and $z = 1$, the coordinates of $A$ are $(-2, 0, 1)$.

1. Express the distance from the origin to $p_1$ in terms of $s$.

The equation of $p_1$ is $2x + sy + z = -3$.
This can be written as $\mathbf{r} \cdot \begin{pmatrix} 2 \\ s \\ 1 \end{pmatrix} = -3$.
The shortest distance from the origin to $p_1$ is:
$\frac{|-3|}{\sqrt{2^2 + s^2 + 1^2}} = \frac{3}{\sqrt{s^2 + 5}}$

2. Equate the expression to the given distance and solve for $s$.

$\frac{3}{\sqrt{s^2 + 5}} = \frac{3}{\sqrt{6}}$
Equating the denominators:
$\sqrt{s^2 + 5} = \sqrt{6}$
$s^2 + 5 = 6$
$s^2 = 1$
$s = \pm 1$
Since it is given in the question that $s$ is a negative real constant, we reject $s = 1$.
$s = -1$

1. Evaluate the position of $B$ relative to plane $p_1$.

The position vector of $B$ is $\mathbf{r}_B = \mathbf{k} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.
For plane $p_1$ ($2x - 2y + z = -3$), its vector equation is $\mathbf{r} \cdot \mathbf{n}_1 = -3$.
$\mathbf{r}_B \cdot \mathbf{n}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} = 1$
Since $1 > -3$, point $B$ lies on the side of $p_1$ that the normal vector $\mathbf{n}_1$ points towards (the direction of increasing dot product). From the diagram, $\mathbf{n}_1$ points into $R_1$ and $R_2$, so $B$ must lie in either $R_1$ or $R_2$.

2. Evaluate the position of $B$ relative to plane $p_2$.

For plane $p_2$ ($x - 2y + 3z = 1$), its vector equation is $\mathbf{r} \cdot \mathbf{n}_2 = 1$.
$\mathbf{r}_B \cdot \mathbf{n}_2 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} = 3$
Since $3 > 1$, point $B$ lies on the side of $p_2$ that the normal vector $\mathbf{n}_2$ points towards. From the diagram, $\mathbf{n}_2$ points into $R_2$ and $R_3$, so $B$ must lie in either $R_2$ or $R_3$.

3. Deduce the final region by finding the intersection of both conditions.

To satisfy both conditions, $B$ must lie in the intersection of the two sets of regions.
$(R_1 \cup R_2) \cap (R_2 \cup R_3) = R_2$
Therefore, point $B$ lies in $R_2$.