平面 $p_1$ 和 $p_2$ 的方程分别为 $\mathbf{r} = \begin{pmatrix} 4 \\ -1 \\ 3 \end{pmatrix} + s \begin{pmatrix} 3 \\ -1 \\ 7 \end{pmatrix} + t \begin{pmatrix} 1 \\ -3 \\ 12 \end{pmatrix}$ 和 $x + 3y = 1$，其中 $s$ 和 $t$ 为参数。

1. Express the general point on $p_1$ in terms of parameters $s$ and $t$, and substitute it into the Cartesian equation of $p_2$.

The parametric equation of $p_1$ gives the general point $(x, y, z)$ as:
$x = 4 + 3s + t$
$y = -1 - s - 3t$
$z = 3 + 7s + 12t$

Substitute $x$ and $y$ into the equation of $p_2$ ($x + 3y = 1$):
$(4 + 3s + t) + 3(-1 - s - 3t) = 1$

2. Solve for the relationship between the parameters $s$ and $t$.

$4 + 3s + t - 3 - 3s - 9t = 1$
$1 - 8t = 1$
$-8t = 0$
$t = 0$

3. Substitute the parameter constraint back into $p_1$ to obtain the vector equation of the line of intersection.

Substitute $t = 0$ back into the vector equation of $p_1$:
$\mathbf{r} = \begin{pmatrix} 4 \\ -1 \\ 3 \end{pmatrix} + s \begin{pmatrix} 3 \\ -1 \\ 7 \end{pmatrix} + (0)\begin{pmatrix} 1 \\ -3 \\ 12 \end{pmatrix}$
$\mathbf{r} = \begin{pmatrix} 4 \\ -1 \\ 3 \end{pmatrix} + s \begin{pmatrix} 3 \\ -1 \\ 7 \end{pmatrix}$

1. Find the normal vector to plane $p_1$.

The normal vector $\mathbf{n}_1$ is the cross product of the two direction vectors of $p_1$:
$\mathbf{n}_1 = \begin{pmatrix} 3 \\ -1 \\ 7 \end{pmatrix} \times \begin{pmatrix} 1 \\ -3 \\ 12 \end{pmatrix} = \begin{pmatrix} (-1)(12) - (7)(-3) \\ (7)(1) - (3)(12) \\ (3)(-3) - (-1)(1) \end{pmatrix} = \begin{pmatrix} 9 \\ -29 \\ -8 \end{pmatrix}$

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

From the Cartesian equation $x + 3y = 1$, the normal vector $\mathbf{n}_2$ is:
$\mathbf{n}_2 = \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix}$

3. Calculate the acute angle between the two planes.

The acute angle $\theta$ between the planes is equal to the acute angle between their normal vectors:
$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$
$\cos \theta = \frac{\left| \begin{pmatrix} 9 \\ -29 \\ -8 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix} \right|}{\sqrt{9^2 + (-29)^2 + (-8)^2}\sqrt{1^2 + 3^2 + 0^2}}$
$\cos \theta = \frac{|9(1) - 29(3) - 8(0)|}{\sqrt{986}\sqrt{10}} = \frac{78}{\sqrt{9860}}$
$\theta = \cos^{-1}\left(\frac{78}{\sqrt{9860}}\right) \approx 38.2^\circ \text{ (1 d.p.)}$

1. Form the equation of the line passing through $A$ and perpendicular to plane $p_2$.

The normal vector to $p_2$ is $\mathbf{n}_2 = \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix}$.
The equation of the line passing through $A$ and perpendicular to $p_2$ is:
$l: \mathbf{r} = \begin{pmatrix} 5 \\ -4 \\ 15 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix}, \quad \lambda \in \mathbb{R}$

2. Substitute the line equation into the plane equation to find the value of $\lambda$ at the intersection.

Since the foot of the perpendicular $F$ lies on $p_2$, we substitute the coordinates of a general point on $l$ into the Cartesian equation of $p_2$ ($x + 3y = 1$):
$(5 + \lambda) + 3(-4 + 3\lambda) = 1$
$5 + \lambda - 12 + 9\lambda = 1$
$-7 + 10\lambda = 1$
$10\lambda = 8 \implies \lambda = \frac{4}{5}$

3. Calculate the position vector of the foot of the perpendicular.

Substitute $\lambda = \frac{4}{5}$ back into the equation of the line:
$\vec{OF} = \begin{pmatrix} 5 \\ -4 \\ 15 \end{pmatrix} + \frac{4}{5} \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix} = \begin{pmatrix} 5 + 0.8 \\ -4 + 2.4 \\ 15 \end{pmatrix} = \begin{pmatrix} 5.8 \\ -1.6 \\ 15 \end{pmatrix}$
Or, in exact fractions: $\begin{pmatrix} \frac{29}{5} \\ -\frac{8}{5} \\ 15 \end{pmatrix}$

1. Find the displacement vector $\vec{AB}$ and its squared magnitude.

$\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix} - \begin{pmatrix} 5 \\ -4 \\ 15 \end{pmatrix} = \begin{pmatrix} -4 \\ 4 \\ -17 \end{pmatrix}$
$|\vec{AB}|^2 = (-4)^2 + 4^2 + (-17)^2 = 16 + 16 + 289 = 321$

2. Calculate the scalar projection of $\vec{AB}$ onto the normal vector of $p_2$.

The normal vector to $p_2$ is $\mathbf{n} = \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix}$.
The scalar projection $d$ of $\vec{AB}$ onto $\mathbf{n}$ is:
$d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{\left| \begin{pmatrix} -4 \\ 4 \\ -17 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix} \right|}{\sqrt{1^2 + 3^2 + 0^2}}$
$d = \frac{|-4(1) + 4(3) + (-17)(0)|}{\sqrt{10}} = \frac{8}{\sqrt{10}}$

3. Use Pythagoras' theorem to find the length of the projection of $AB$ onto $p_2$.

Let $L$ be the length of the projection of $AB$ onto $p_2$. Using Pythagoras' theorem on the right-angled triangle formed by $\vec{AB}$, its projection on the plane, and its projection on the normal:
$L = \sqrt{|\vec{AB}|^2 - d^2}$
$L = \sqrt{321 - \left(\frac{8}{\sqrt{10}}\right)^2} = \sqrt{321 - 6.4}$
$L = \sqrt{314.6} \approx 17.7 \text{ (to 3 s.f.)}$