以点 $O$ 为原点，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 $PR$ and formulate its vector equation.

$\vec{PR} = \vec{OR} - \vec{OP} = \begin{pmatrix} 5 \\ -2 \\ -1 \end{pmatrix} - \begin{pmatrix} -3 \\ 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 8 \\ -6 \\ -4 \end{pmatrix}$
Taking out the common factor of 2, a parallel direction vector is $\mathbf{d} = \begin{pmatrix} 4 \\ -3 \\ -2 \end{pmatrix}$.
The line $PR$ has the equation $\mathbf{r} = \begin{pmatrix} -3 \\ 4 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 4 \\ -3 \\ -2 \end{pmatrix}$, where $\lambda \in \mathbb{R}$.

2. Express the general point $S$ on the line and apply the condition for minimum distance.

For the distance from $O$ to $S$ to be a minimum, $\vec{OS}$ must be perpendicular to the line $PR$. Hence, $\vec{OS} \cdot \mathbf{d} = 0$.
Since $S$ lies on $PR$, $\vec{OS} = \begin{pmatrix} -3 + 4\lambda \\ 4 - 3\lambda \\ 3 - 2\lambda \end{pmatrix}$ for some $\lambda$.
$\begin{pmatrix} -3 + 4\lambda \\ 4 - 3\lambda \\ 3 - 2\lambda \end{pmatrix} \cdot \begin{pmatrix} 4 \\ -3 \\ -2 \end{pmatrix} = 0$
$4(-3 + 4\lambda) - 3(4 - 3\lambda) - 2(3 - 2\lambda) = 0$
$-12 + 16\lambda - 12 + 9\lambda - 6 + 4\lambda = 0$
$29\lambda - 30 = 0 \implies \lambda = \frac{30}{29}$

3. Substitute $\lambda$ back into $\vec{OS}$ to find the coordinates of $S$.

$\vec{OS} = \begin{pmatrix} -3 + 4\left(\frac{30}{29}\right) \\ 4 - 3\left(\frac{30}{29}\right) \\ 3 - 2\left(\frac{30}{29}\right) \end{pmatrix} = \begin{pmatrix} -3 + \frac{120}{29} \\ 4 - \frac{90}{29} \\ 3 - \frac{60}{29} \end{pmatrix} = \begin{pmatrix} \frac{33}{29} \\ \frac{26}{29} \\ \frac{27}{29} \end{pmatrix}$
The coordinates of $S$ are $\left(\frac{33}{29}, \frac{26}{29}, \frac{27}{29}\right)$.

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

The normal vector $\mathbf{n}$ to the plane $OPQR$ is perpendicular to both $\vec{OP}$ and $\vec{OR}$. We can find it using their cross product:
$\mathbf{n} = \vec{OP} \times \vec{OR} = \begin{pmatrix} -3 \\ 4 \\ 3 \end{pmatrix} \times \begin{pmatrix} 5 \\ -2 \\ -1 \end{pmatrix}$
$\mathbf{n} = \begin{pmatrix} (4)(-1) - (3)(-2) \\ (3)(5) - (-3)(-1) \\ (-3)(-2) - (4)(5) \end{pmatrix} = \begin{pmatrix} 2 \\ 12 \\ -14 \end{pmatrix}$
We can use the simplified normal vector $\mathbf{n_1} = \begin{pmatrix} 1 \\ 6 \\ -7 \end{pmatrix}$.

2. Formulate the distance equation for the parallel planes.

Since the required planes are parallel to the base $OPQR$, their cartesian equations take the form $x + 6y - 7z = d$.

The base $OPQR$ passes through the origin $(0, 0, 0)$. The perpendicular distance from a plane to the base is therefore the perpendicular distance from the origin to the plane, which is given as $2\sqrt{86}$.
$\frac{|(1)(0) + (6)(0) - (7)(0) - d|}{\sqrt{1^2 + 6^2 + (-7)^2}} = 2\sqrt{86}$
$\frac{|-d|}{\sqrt{1 + 36 + 49}} = 2\sqrt{86}$
$\frac{|d|}{\sqrt{86}} = 2\sqrt{86}$

3. Solve for the constants and state the cartesian equations of the planes.

Solving for $d$:
$|d| = 2(\sqrt{86})^2 = 2(86) = 172$
So $d = 172$ or $d = -172$.

Therefore, the cartesian equations of the planes are:
$x + 6y - 7z = 172$ and $x + 6y - 7z = -172$

1. Find the direction vector of $OV$ by identifying it as the normal to the base $OPQR$.

Since $OV$ is the height of the pyramid, the line segment $OV$ is perpendicular to the base $OPQR$. The direction of $OV$ is parallel to the normal vector of the base:
$\mathbf{n} = \vec{OP} \times \vec{OR} = \begin{pmatrix} -3 \\ 4 \\ 3 \end{pmatrix} \times \begin{pmatrix} 5 \\ -2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ 12 \\ -14 \end{pmatrix}$

2. Calculate the acute angle between $OV$ and the vertical using the dot product.

The horizontal plane is the $x$-$y$ plane, so the vertical direction is parallel to the $z$-axis, represented by the vector $\mathbf{k} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$. Let $\theta$ be the acute angle between $OV$ and the vertical:
$\cos \theta = \frac{|\mathbf{n} \cdot \mathbf{k}|}{|\mathbf{n}||\mathbf{k}|}$
$\cos \theta = \frac{\left| \begin{pmatrix} 2 \\ 12 \\ -14 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right|}{\sqrt{2^2 + 12^2 + (-14)^2} (1)} = \frac{|-14|}{\sqrt{344}} = \frac{14}{2\sqrt{86}} = \frac{7}{\sqrt{86}}$
$\theta = \cos^{-1}\left(\frac{7}{\sqrt{86}}\right) \approx 40.985^\circ$
The acute angle is $41.0^\circ$ (to 1 d.p.).

1. Relate the given height $OV$ to the normal of the base and find an expression for $\vec{OV}$.

Since the height of the pyramid is given by the exact length of the segment $OV$ and $O$ is a point on the base, $OV$ must be perpendicular to the base $OPQR$.
Therefore, $\vec{OV}$ is parallel to the normal vector of the plane containing the base, which is given by $\vec{OP} \times \vec{OR}$.
$\vec{OP} \times \vec{OR} = \begin{pmatrix} -3 \\ 4 \\ 3 \end{pmatrix} \times \begin{pmatrix} 5 \\ -2 \\ -1 \end{pmatrix} = \begin{pmatrix} (4)(-1) - (3)(-2) \\ (3)(5) - (-3)(-1) \\ (-3)(-2) - (4)(5) \end{pmatrix} = \begin{pmatrix} 2 \\ 12 \\ -14 \end{pmatrix}$
Let $\vec{OV} = \mu \begin{pmatrix} 2 \\ 12 \\ -14 \end{pmatrix}$ for some scalar $\mu$.

2. Find the position vector of the fourth vertex $Q$ of the parallelogram base.

The vertices of the parallelogram base in order are $O, P, Q, R$. Thus, the diagonal vector $\vec{OQ}$ is the sum of the adjacent side vectors $\vec{OP}$ and $\vec{OR}$:
$\vec{OQ} = \vec{OP} + \vec{OR} = \begin{pmatrix} -3 \\ 4 \\ 3 \end{pmatrix} + \begin{pmatrix} 5 \\ -2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix}$

3. Express $\vec{QV}$ and solve for $\mu$ using the given parallel condition to find $\vec{OV}$.

$\vec{QV} = \vec{OV} - \vec{OQ} = \begin{pmatrix} 2\mu \\ 12\mu \\ -14\mu \end{pmatrix} - \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 2\mu - 2 \\ 12\mu - 2 \\ -14\mu - 2 \end{pmatrix}$
We are given that $\vec{QV}$ is parallel to $-5\mathbf{j} + 8\mathbf{k} = \begin{pmatrix} 0 \\ -5 \\ 8 \end{pmatrix}$.
For the vectors to be parallel, the $x$-component of $\vec{QV}$ must be zero:
$2\mu - 2 = 0 \implies \mu = 1$
When $\mu = 1$, $\vec{QV} = \begin{pmatrix} 0 \\ 10 \\ -16 \end{pmatrix} = -2\begin{pmatrix} 0 \\ -5 \\ 8 \end{pmatrix}$, verifying that they are indeed parallel.
Thus, substituting $\mu = 1$, we get:
$\vec{OV} = \begin{pmatrix} 2 \\ 12 \\ -14 \end{pmatrix} = 2\mathbf{i} + 12\mathbf{j} - 14\mathbf{k}$

4. Calculate the exact volume of the pyramid.

The base area is the area of the parallelogram $OPQR$:
$\text{Base Area} = |\vec{OP} \times \vec{OR}| = \left| \begin{pmatrix} 2 \\ 12 \\ -14 \end{pmatrix} \right| = \sqrt{2^2 + 12^2 + (-14)^2} = \sqrt{4 + 144 + 196} = \sqrt{344}$
The height of the pyramid is exactly the length of $OV$:
$\text{Height} = |\vec{OV}| = \left| \begin{pmatrix} 2 \\ 12 \\ -14 \end{pmatrix} \right| = \sqrt{344}$
Using the given formula for the volume of a pyramid:
$\text{Volume} = \frac{1}{3} \times \text{Base Area} \times \text{Height} = \frac{1}{3}(\sqrt{344})(\sqrt{344}) = \frac{344}{3} \text{ (or } 114\frac{2}{3}\text{)}$