直线 $l$ 经过坐标为 $(1, -5, -5)$ 的点 $P$，且平行于 $3\mathbf{i} + \mathbf{j} - 6\mathbf{k}$。平面 $\pi_1$ 的笛卡尔方程为 $2x - 5y - 3z = 4$。

1. Identify the direction vector of the line and the normal vector of the plane.

Let the direction vector of $l$ be $\mathbf{d} = \begin{pmatrix} 3 \\ 1 \\ -6 \end{pmatrix}$.
Let the normal vector of $\pi_1$ be $\mathbf{n} = \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix}$.

2. Calculate the scalar product and the magnitudes of the direction and normal vectors.

$\mathbf{d} \cdot \mathbf{n} = \begin{pmatrix} 3 \\ 1 \\ -6 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} = (3)(2) + (1)(-5) + (-6)(-3) = 6 - 5 + 18 = 19$
$|\mathbf{d}| = \sqrt{3^2 + 1^2 + (-6)^2} = \sqrt{9 + 1 + 36} = \sqrt{46}$
$|\mathbf{n}| = \sqrt{2^2 + (-5)^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$

3. Use the sine formula to compute the acute angle between the line and the plane.

Let $\theta$ be the acute angle between $l$ and $\pi_1$.
$\sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}| |\mathbf{n}|}$
$\sin \theta = \frac{|19|}{\sqrt{46}\sqrt{38}} = \frac{19}{\sqrt{1748}}$
$\theta = \sin^{-1}\left(\frac{19}{\sqrt{1748}}\right) \approx 27.031^\circ$
Rounding to the nearest degree, the angle is $27^\circ$.

1. Write down the parametric or vector equation of line $l$.

The line $l$ passes through $(1, -5, -5)$ and is parallel to $3\mathbf{i} + \mathbf{j} - 6\mathbf{k}$. Its vector equation is:
$\mathbf{r} = \begin{pmatrix} 1 \\ -5 \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 1 \\ -6 \end{pmatrix}, \text{ for } \lambda \in \mathbb{R}.$
Thus, any point on the line has coordinates $(1 + 3\lambda, -5 + \lambda, -5 - 6\lambda)$.

2. Substitute the general point on $l$ into the plane's cartesian equation to solve for the parameter $\lambda$.

Substitute $x = 1 + 3\lambda$, $y = -5 + \lambda$, and $z = -5 - 6\lambda$ into the equation of $\pi_1$, $2x - 5y - 3z = 4$:
$2(1 + 3\lambda) - 5(-5 + \lambda) - 3(-5 - 6\lambda) = 4$
$2 + 6\lambda + 25 - 5\lambda + 15 + 18\lambda = 4$
$19\lambda + 42 = 4$
$19\lambda = -38$
$\lambda = -2$

3. Calculate the coordinates of the intersection point $Q$.

Substitute $\lambda = -2$ back into the line equation:
$\mathbf{r}_Q = \begin{pmatrix} 1 \\ -5 \\ -5 \end{pmatrix} - 2 \begin{pmatrix} 3 \\ 1 \\ -6 \end{pmatrix} = \begin{pmatrix} 1 - 6 \\ -5 - 2 \\ -5 + 12 \end{pmatrix} = \begin{pmatrix} -5 \\ -7 \\ 7 \end{pmatrix}$
The coordinates of $Q$ are $(-5, -7, 7)$.

1. Identify two vectors parallel to the plane $\pi_2$.

Since $\pi_2$ contains line $l$, it is parallel to the direction vector of $l$, $\mathbf{d} = \begin{pmatrix} 3 \\ 1 \\ -6 \end{pmatrix}$.
Since $\pi_2$ is perpendicular to $\pi_1$, it must be parallel to the normal vector of $\pi_1$, which is $\mathbf{n}_1 = \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix}$.

2. Calculate the normal vector of $\pi_2$ using the cross product.

The normal vector to $\pi_2$ is perpendicular to both $\mathbf{d}$ and $\mathbf{n}_1$:
$\mathbf{n}_2 = \mathbf{d} \times \mathbf{n}_1 = \begin{pmatrix} 3 \\ 1 \\ -6 \end{pmatrix} \times \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} = \begin{pmatrix} 1(-3) - (-6)(-5) \\ -6(2) - 3(-3) \\ 3(-5) - 1(2) \end{pmatrix} = \begin{pmatrix} -33 \\ -3 \\ -17 \end{pmatrix}$
To simplify the Cartesian equation, we can take the normal vector as the parallel vector $\begin{pmatrix} 33 \\ 3 \\ 17 \end{pmatrix}$.

3. Use point $P$ to formulate the Cartesian equation of $\pi_2$.

Point $P(1, -5, -5)$ lies on $l$ and thus on $\pi_2$. Using the scalar product form $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$:
$\mathbf{a} \cdot \mathbf{n}_2 = \begin{pmatrix} 1 \\ -5 \\ -5 \end{pmatrix} \cdot \begin{pmatrix} 33 \\ 3 \\ 17 \end{pmatrix} = 33 - 15 - 85 = -67$
Therefore, the Cartesian equation of $\pi_2$ is $33x + 3y + 17z = -67$.

1. Determine the point of intersection $Q$ of line $l$ and plane $\pi_1$ (requires result from part ii).

$l: \mathbf{r} = \begin{pmatrix} 1 \\ -5 \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 1 \\ -6 \end{pmatrix}$
Substitute into $\pi_1: 2x - 5y - 3z = 4$:
$\begin{pmatrix} 1+3\lambda \\ -5+\lambda \\ -5-6\lambda \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} = 4$
$2(1+3\lambda) - 5(-5+\lambda) - 3(-5-6\lambda) = 4$
$2 + 6\lambda + 25 - 5\lambda + 15 + 18\lambda = 4$
$19\lambda = -38 \implies \lambda = -2$
Substitute $\lambda = -2$ to find the position vector of $Q$:
$\vec{OQ} = \begin{pmatrix} 1 \\ -5 \\ -5 \end{pmatrix} - 2 \begin{pmatrix} 3 \\ 1 \\ -6 \end{pmatrix} = \begin{pmatrix} -5 \\ -7 \\ 7 \end{pmatrix}$

2. Determine the normal vector of plane $\pi_2$ (requires result from part iii).

Since $\pi_2$ contains $l$ and is perpendicular to $\pi_1$, its normal $\mathbf{n}_2$ is perpendicular to both the direction of $l$ and the normal of $\pi_1$.
$\mathbf{n}_2 = \begin{pmatrix} 3 \\ 1 \\ -6 \end{pmatrix} \times \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} = \begin{pmatrix} (1)(-3) - (-6)(-5) \\ (-6)(2) - (3)(-3) \\ (3)(-5) - (1)(2) \end{pmatrix} = \begin{pmatrix} -33 \\ -3 \\ -17 \end{pmatrix}$
We can take the normal vector to be $\mathbf{n}_2 = \begin{pmatrix} 33 \\ 3 \\ 17 \end{pmatrix}$.

3. Find the direction vector of the line of intersection between $\pi_1$ and $\pi_2$.

The line of intersection lies on both planes, so its direction vector $\mathbf{d}$ is perpendicular to both $\mathbf{n}_1$ and $\mathbf{n}_2$:
$\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} \times \begin{pmatrix} 33 \\ 3 \\ 17 \end{pmatrix}$
$\mathbf{d} = \begin{pmatrix} (-5)(17) - (-3)(3) \\ (-3)(33) - (2)(17) \\ (2)(3) - (-5)(33) \end{pmatrix} = \begin{pmatrix} -85 + 9 \\ -99 - 34 \\ 6 + 165 \end{pmatrix} = \begin{pmatrix} -76 \\ -133 \\ 171 \end{pmatrix}$
Factoring out $-19$, we get a simpler parallel direction vector:
$\mathbf{d} = -19 \begin{pmatrix} 4 \\ 7 \\ -9 \end{pmatrix}$
We use $\begin{pmatrix} 4 \\ 7 \\ -9 \end{pmatrix}$ as the direction vector for the line.

4. Formulate the vector equation of the intersection line.

The line of intersection must pass through $Q$, which is the intersection of $l$ (which lies in $\pi_2$) and $\pi_1$. Thus, $Q(-5, -7, 7)$ is a point on the line.
Using the point and direction vector:
$\mathbf{r} = \begin{pmatrix} -5 \\ -7 \\ 7 \end{pmatrix} + \mu \begin{pmatrix} 4 \\ 7 \\ -9 \end{pmatrix}, \text{ where } \mu \in \mathbb{R}$