以点 $O$ 为原点且以 $x$-$y$ 平面为水平面，棱锥 $OPQRV$ 具有平行四边形底面 $OPQR$ 和高 $OV$。点 $P$ 和 $R$ 的位置向量分别为 $-3\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}$ 和 $5\mathbf{i} - 2\mathbf{j} - \mathbf{k}$。

1. Find the direction vector of the line $PR$.

$\overrightarrow{PR} = \overrightarrow{OR} - \overrightarrow{OP} = \begin{pmatrix} 5 \\ -2 \\ -1 \end{pmatrix} - \begin{pmatrix} -3 \\ 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 8 \\ -6 \\ -4 \end{pmatrix}$
We can use a simpler direction vector $\mathbf{d}$ parallel to $\overrightarrow{PR}$ by dividing by $2$:
$\mathbf{d} = \begin{pmatrix} 4 \\ -3 \\ -2 \end{pmatrix}$

2. Set up the position vector of $S$ and apply the perpendicularity condition for the minimum distance.

Since $S$ lies on the line $PR$, its position vector can be expressed in terms of a parameter $\lambda$:
$\overrightarrow{OS} = \overrightarrow{OP} + \lambda \mathbf{d} = \begin{pmatrix} -3 \\ 4 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ -3 \\ -2 \end{pmatrix} = \begin{pmatrix} -3 + 4\lambda \\ 4 - 3\lambda \\ 3 - 2\lambda \end{pmatrix}$
For the distance from $O$ to $S$ to be a minimum, $\overrightarrow{OS}$ must be perpendicular to the line $PR$. Therefore, their dot product is zero:
$\overrightarrow{OS} \cdot \mathbf{d} = 0$
$\begin{pmatrix} -3 + 4\lambda \\ 4 - 3\lambda \\ 3 - 2\lambda \end{pmatrix} \cdot \begin{pmatrix} 4 \\ -3 \\ -2 \end{pmatrix} = 0$
$4(-3 + 4\lambda) - 3(4 - 3\lambda) - 2(3 - 2\lambda) = 0$

3. Solve for the parameter $\lambda$ and find the coordinates of $S$.

$-12 + 16\lambda - 12 + 9\lambda - 6 + 4\lambda = 0$
$29\lambda - 30 = 0 \implies \lambda = \frac{30}{29}$
Substitute $\lambda = \frac{30}{29}$ back into $\overrightarrow{OS}$:
$\overrightarrow{OS} = \begin{pmatrix} -3 + 4\left(\frac{30}{29}\right) \\ 4 - 3\left(\frac{30}{29}\right) \\ 3 - 2\left(\frac{30}{29}\right) \end{pmatrix} = \begin{pmatrix} \frac{-87+120}{29} \\ \frac{116-90}{29} \\ \frac{87-60}{29} \end{pmatrix} = \begin{pmatrix} \frac{33}{29} \\ \frac{26}{29} \\ \frac{27}{29} \end{pmatrix}$
Thus, the coordinates of $S$ are $\left( \frac{33}{29}, \frac{26}{29}, \frac{27}{29} \right)$.

1. Find the normal vector to the base plane $OPQR$.

The base plane $OPQR$ contains the vectors $\vec{OP}$ and $\vec{OR}$. Its normal vector $\mathbf{n}$ can be found using the cross product:
$\mathbf{n} = \vec{OP} \times \vec{OR} = \begin{pmatrix} -3 \\ 4 \\ 3 \end{pmatrix} \times \begin{pmatrix} 5 \\ -2 \\ -1 \end{pmatrix}$
$= \begin{pmatrix} (4)(-1) - (3)(-2) \\ (3)(5) - (-3)(-1) \\ (-3)(-2) - (4)(5) \end{pmatrix} = \begin{pmatrix} -4 + 6 \\ 15 - 3 \\ 6 - 20 \end{pmatrix} = \begin{pmatrix} 2 \\ 12 \\ -14 \end{pmatrix}$
We can take a simpler normal vector by dividing by 2: $\mathbf{n}' = \begin{pmatrix} 1 \\ 6 \\ -7 \end{pmatrix}$.

2. Set up an equation for the parallel planes using the given perpendicular distance.

Since the required planes are parallel to the base, they share the same normal direction. Let their cartesian equation be of the form $x + 6y - 7z = d$.

The base plane passes through the origin $(0,0,0)$, so its equation is $x + 6y - 7z = 0$.

The perpendicular distance from the origin to a parallel plane is given by $\frac{|d|}{\sqrt{1^2 + 6^2 + (-7)^2}}$.
Equating this to the given distance of $2\sqrt{86}$:
$\frac{|d|}{\sqrt{1 + 36 + 49}} = 2\sqrt{86}$
$\frac{|d|}{\sqrt{86}} = 2\sqrt{86}$

3. Solve for the unknown constant $d$ and state the equations.

$|d| = 2(\sqrt{86})^2 = 2(86) = 172$
$d = 172 \quad \text{or} \quad d = -172$

The cartesian equations of the planes are $x + 6y - 7z = 172$ and $x + 6y - 7z = -172$.

1. Identify the direction vectors for $OV$ and the vertical.

Since the $x$-$y$ plane is given as a horizontal plane, the vertical direction is parallel to the $z$-axis, which has the unit direction vector $\mathbf{k} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.

The line segment $OV$ is the height of the pyramid, which implies that it is perpendicular to the base $OPQR$. The direction of $OV$ is therefore parallel to the normal vector $\mathbf{n}$ of the base. We can find $\mathbf{n}$ by taking the cross product of two vectors on the base:
$\mathbf{n} = \overrightarrow{OP} \times \overrightarrow{OR} = \begin{pmatrix} -3 \\ 4 \\ 3 \end{pmatrix} \times \begin{pmatrix} 5 \\ -2 \\ -1 \end{pmatrix}$
$\mathbf{n} = \begin{pmatrix} 4(-1) - 3(-2) \\ 3(5) - (-3)(-1) \\ (-3)(-2) - 4(5) \end{pmatrix} = \begin{pmatrix} 2 \\ 12 \\ -14 \end{pmatrix} = 2\begin{pmatrix} 1 \\ 6 \\ -7 \end{pmatrix}$
The direction vector of $OV$ can be taken as $\mathbf{d} = \begin{pmatrix} 1 \\ 6 \\ -7 \end{pmatrix}$.

2. Calculate the acute angle between the two direction vectors.

Let $\theta$ be the acute angle between $OV$ and the vertical. Using the scalar product:
$\cos \theta = \frac{|\mathbf{d} \cdot \mathbf{k}|}{|\mathbf{d}| |\mathbf{k}|}$
$\cos \theta = \frac{\left| \begin{pmatrix} 1 \\ 6 \\ -7 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right|}{\sqrt{1^2 + 6^2 + (-7)^2} \sqrt{0^2 + 0^2 + 1^2}}$
$\cos \theta = \frac{|-7|}{\sqrt{1 + 36 + 49} (1)} = \frac{7}{\sqrt{86}}$
$\theta = \cos^{-1}\left(\frac{7}{\sqrt{86}}\right) \approx 40.999^{\circ}$
$\theta = 41.0^{\circ}$ (to 1 d.p.)

1. Find the normal vector to the base $OPQR$ to determine the direction of $\overrightarrow{OV}$.

Since $OV$ is the height of the pyramid, the segment $OV$ is perpendicular to the base $OPQR$. The direction of $\overrightarrow{OV}$ is thus parallel to the normal vector of the base, which is given by the cross product $\overrightarrow{OP} \times \overrightarrow{OR}$.
$\mathbf{n} = \overrightarrow{OP} \times \overrightarrow{OR} = \begin{pmatrix} -3 \\ 4 \\ 3 \end{pmatrix} \times \begin{pmatrix} 5 \\ -2 \\ -1 \end{pmatrix} = \begin{pmatrix} (4)(-1) - (3)(-2) \\ (3)(5) - (-3)(-1) \\ (-3)(-2) - (4)(5) \end{pmatrix} = \begin{pmatrix} 2 \\ 12 \\ -14 \end{pmatrix}$
Let $\overrightarrow{OV} = k \begin{pmatrix} 1 \\ 6 \\ -7 \end{pmatrix}$ for some scalar $k$.

2. Express $\overrightarrow{OQ}$ using vector addition.

Since the base $OPQR$ is a parallelogram, the position vector of $Q$ is the sum of the position vectors of $P$ and $R$:
$\overrightarrow{OQ} = \overrightarrow{OP} + \overrightarrow{OR} = \begin{pmatrix} -3 \\ 4 \\ 3 \end{pmatrix} + \begin{pmatrix} 5 \\ -2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix}$

3. Use the parallel condition to find the scalar $k$ and the position vector of $V$.

We find $\overrightarrow{QV}$ in terms of $k$:
$\overrightarrow{QV} = \overrightarrow{OV} - \overrightarrow{OQ} = \begin{pmatrix} k \\ 6k \\ -7k \end{pmatrix} - \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} k - 2 \\ 6k - 2 \\ -7k - 2 \end{pmatrix}$
We are given that $\overrightarrow{QV}$ is parallel to $-5\mathbf{j} + 8\mathbf{k} = \begin{pmatrix} 0 \\ -5 \\ 8 \end{pmatrix}$. This implies that the $\mathbf{i}$-component of $\overrightarrow{QV}$ must be $0$:
$k - 2 = 0 \implies k = 2$
Substituting $k = 2$ gives the position vector of $V$:
$\overrightarrow{OV} = \begin{pmatrix} 2 \\ 12 \\ -14 \end{pmatrix} = 2\mathbf{i} + 12\mathbf{j} - 14\mathbf{k}$

4. Calculate the base area and the height of the pyramid.

The base area is the magnitude of the cross product of the vectors forming the parallelogram:
$\text{Base area} = |\overrightarrow{OP} \times \overrightarrow{OR}| = \left| \begin{pmatrix} 2 \\ 12 \\ -14 \end{pmatrix} \right| = \sqrt{2^2 + 12^2 + (-14)^2} = \sqrt{344} = 2\sqrt{86}$
The height of the pyramid is the length of the segment $OV$:
$\text{Height} = |\overrightarrow{OV}| = \left| \begin{pmatrix} 2 \\ 12 \\ -14 \end{pmatrix} \right| = 2\sqrt{86}$

5. Find the exact volume of the pyramid.

Using the given formula for the volume of a pyramid:
$\text{Volume} = \frac{1}{3} \times \text{Base area} \times \text{Height}$
$\text{Volume} = \frac{1}{3} \left( 2\sqrt{86} \right) \left( 2\sqrt{86} \right) = \frac{4(86)}{3} = \frac{344}{3}$