已知 $a$ 和 $b$ 为正实数，直线 $l_1$ 和 $l_2$ 的方程由下式给出
$l_1: \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix} \text{ for } \lambda \in \mathbb{R} \quad \text{and} \quad l_2: \mathbf{r} = \begin{pmatrix} a \\ 9 \\ 9 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 7 \\ b \end{pmatrix} \text{ for } \mu \in \mathbb{R}.$
已知这两条直线相交。

1. Equate the position vectors of $l_1$ and $l_2$ to form a system of equations.

Since the lines intersect, there exist parameters $\lambda$ and $\mu$ such that:
$\begin{pmatrix} 1 - \lambda \\ 2 \\ 5 + 3\lambda \end{pmatrix} = \begin{pmatrix} a + \mu \\ 9 + 7\mu \\ 9 + b\mu \end{pmatrix}$

Comparing components, we get:
(1) $1 - \lambda = a + \mu$
(2) $2 = 9 + 7\mu$
(3) $5 + 3\lambda = 9 + b\mu$

2. Solve for $\mu$ and express $\lambda$ in terms of $a$.

From equation (2):
$7\mu = -7 \implies \mu = -1$

Substitute $\mu = -1$ into equation (1):
$1 - \lambda = a - 1$
$\lambda = 2 - a$

3. Substitute the parameters into the third equation to show the required result.

Substitute $\mu = -1$ and $\lambda = 2 - a$ into equation (3):
$5 + 3(2 - a) = 9 + b(-1)$
$5 + 6 - 3a = 9 - b$
$11 - 3a = 9 - b$
$3a = b + 2 \text{ (shown)}$

1. State the direction vectors and set up the equation for the angle between the two lines.

The direction vectors are $\mathbf{d}_1 = \begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix}$ and $\mathbf{d}_2 = \begin{pmatrix} 1 \\ 7 \\ b \end{pmatrix}$.
The angle $\theta$ between the lines satisfies $\cos \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1| |\mathbf{d}_2|}$.
So, $\frac{\left|\begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 7 \\ b \end{pmatrix}\right|}{\left|\begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix}\right| \left|\begin{pmatrix} 1 \\ 7 \\ b \end{pmatrix}\right|} = \frac{11}{\sqrt{660}}$

2. Evaluate the scalar product and magnitudes, then simplify the equation.

$\frac{|(-1)(1) + (0)(7) + (3)(b)|}{\sqrt{(-1)^2 + 0^2 + 3^2} \sqrt{1^2 + 7^2 + b^2}} = \frac{11}{\sqrt{660}}$
$\frac{|3b - 1|}{\sqrt{10} \sqrt{50 + b^2}} = \frac{11}{\sqrt{660}}$

3. Square both sides to form a quadratic equation in $b$.

Squaring both sides gives:
$\frac{(3b - 1)^2}{10(50 + b^2)} = \frac{121}{660}$
$\frac{9b^2 - 6b + 1}{50 + b^2} = \frac{121}{66} = \frac{11}{6}$
$6(9b^2 - 6b + 1) = 11(50 + b^2)$
$54b^2 - 36b + 6 = 550 + 11b^2$
$43b^2 - 36b - 544 = 0$

4. Find the roots of the quadratic equation and state the valid value for $b$.

Solving the quadratic equation:
$(b - 4)(43b + 136) = 0$
$b = 4 \text{ or } b = -\frac{136}{43}$
Since it is given that $a$ and $b$ are positive real numbers, we reject the negative root.
$b = 4$

1. Find a second direction vector parallel to the plane $\Pi_2$.

Let $A$ be the point $(1, 2, 5)$ on $l_1$ and $P$ be the point $(2, 1, 3)$.

$\vec{AP} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix}$

2. Find the normal vector to the plane $\Pi_2$ using the cross product.

The normal vector $\mathbf{n}_2$ is perpendicular to both $\vec{AP}$ and the direction vector of $l_1$, $\begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix}$.

$\mathbf{n}_2 = \begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix} = \begin{pmatrix} 0(-2) - 3(-1) \\ 3(1) - (-1)(-2) \\ (-1)(-1) - 0(1) \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}$

3. Form the scalar product equation of the plane.

Using the point $A(1, 2, 5)$ which lies on $\Pi_2$, the equation is $\mathbf{r} \cdot \mathbf{n}_2 = \mathbf{a} \cdot \mathbf{n}_2$.

$\mathbf{r} \cdot \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} = 3(1) + 1(2) + 1(5)$
$\mathbf{r} \cdot \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} = 10$

1. State the normal vector of $\Pi_1$ and calculate the normal vector of $\Pi_2$ (required from part iii).

From the given equation of $\Pi_1$, its normal vector is $\mathbf{n}_1 = \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix}$.

For $\Pi_2$, which contains line $l_1$ and the point $A(2, 1, 3)$, we first identify a point on $l_1$, $B(1, 2, 5)$, and the line's direction vector $\mathbf{d}_1 = \begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix}$.
The vector connecting $B$ to $A$ is $\vec{BA} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix}$.
The normal to $\Pi_2$ is found using the vector product:
$\mathbf{n}_2 = \mathbf{d}_1 \times \vec{BA} = \begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix} = \begin{pmatrix} 0(-2) - 3(-1) \\ 3(1) - (-1)(-2) \\ (-1)(-1) - 0(1) \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}$

2. Calculate the acute angle between the two planes using their normal vectors.

Let $\theta$ be the angle between the planes, which corresponds to the acute angle between their normal vectors:
$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| |\mathbf{n}_2|}$
$\cos \theta = \frac{\left| \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} \right|}{\sqrt{3^2 + (-1)^2 + 1^2} \sqrt{3^2 + 1^2 + 1^2}}$
$\cos \theta = \frac{|(3)(3) + (-1)(1) + (1)(1)|}{\sqrt{11} \sqrt{11}}$
$\cos \theta = \frac{9}{11}$
$\theta = \cos^{-1}\left(\frac{9}{11}\right) \approx 35.1^\circ$

1. State the general equation of a plane parallel to $\Pi_1$.

Since $\Pi_3$ is parallel to $\Pi_1$, they share the same normal vector $\mathbf{n} = \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix}$.
Let the equation of $\Pi_3$ be $\mathbf{r} \cdot \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix} = d$.

2. Use the distance between the two planes to form an equation in $d$.

The perpendicular distance between the parallel planes $\mathbf{r} \cdot \mathbf{n} = d$ and $\mathbf{r} \cdot \mathbf{n} = 6$ is given by $\frac{|6 - d|}{|\mathbf{n}|}$.
$|\mathbf{n}| = \sqrt{3^2 + (-1)^2 + 1^2} = \sqrt{11}$
$\frac{|6 - d|}{\sqrt{11}} = \frac{7}{\sqrt{11}}$
$|6 - d| = 7$
$6 - d = 7 \implies d = -1 \quad \text{or} \quad 6 - d = -7 \implies d = 13$

3. Determine the correct value of $d$ using the condition that $\Pi_3$ is closer to the origin, and express in cartesian form.

The perpendicular distance from the origin to a plane $\mathbf{r} \cdot \mathbf{n} = D$ is $\frac{|D|}{|\mathbf{n}|}$.
For $\Pi_1$, the distance to the origin is $\frac{6}{\sqrt{11}}$.
For $\Pi_3$, the distance to the origin is $\frac{|-1|}{\sqrt{11}} = \frac{1}{\sqrt{11}}$ or $\frac{|13|}{\sqrt{11}} = \frac{13}{\sqrt{11}}$.
Since $\Pi_3$ is closer to the origin than $\Pi_1$, we must have $\frac{|d|}{\sqrt{11}} < \frac{6}{\sqrt{11}}$, which implies $d = -1$.
The scalar product form is $\mathbf{r} \cdot \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix} = -1$.
Converting to cartesian form, we obtain:
$3x - y + z = -1$