直线 $l$ 的方程为 $\mathbf{r} = i + 6j - 3k + s(i - 2j + 2k)。$ 平面 $p$ 的方程为 $(\mathbf{r} - 3i).(2i - 3j + 6k) = 0。$ 直线 $l$ 与平面 $p$ 相交于点 $A$。

1. Express the equation of the plane in the standard scalar product form $\mathbf{r} \cdot \mathbf{n} = d$.

$(\mathbf{r} - 3\mathbf{i}) \cdot (2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) = 0$
$\mathbf{r} \cdot (2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) = 3\mathbf{i} \cdot (2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k})$
$\mathbf{r} \cdot (2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) = (3)(2) + (0)(-3) + (0)(6) = 6$

2. Substitute the equation of the line into the equation of the plane to find the parameter $s$ at the point of intersection.

Substitute $\mathbf{r} = (1+s)\mathbf{i} + (6-2s)\mathbf{j} + (-3+2s)\mathbf{k}$ into the plane equation:
$[(1+s)\mathbf{i} + (6-2s)\mathbf{j} + (-3+2s)\mathbf{k}] \cdot (2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) = 6$
$2(1+s) - 3(6-2s) + 6(-3+2s) = 6$
$2 + 2s - 18 + 6s - 18 + 12s = 6$
$20s - 34 = 6$
$20s = 40 \implies s = 2$

3. Substitute the value of $s$ back into the line equation to obtain the position vector of point $A$.

$\mathbf{r}_A = \mathbf{i} + 6\mathbf{j} - 3\mathbf{k} + 2(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k})$
$\mathbf{r}_A = (1+2)\mathbf{i} + (6-4)\mathbf{j} + (-3+4)\mathbf{k}$
$\mathbf{r}_A = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k}$

1. Identify the direction vector of the line $l$ and the normal vector of the plane $p$.

The direction vector of line $l$ is $\mathbf{d} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$.

The normal vector of plane $p$ is $\mathbf{n} = \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}$.

2. Calculate the scalar product of the direction vector and the normal vector.

$\mathbf{d} \cdot \mathbf{n} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}$
$\mathbf{d} \cdot \mathbf{n} = (1)(2) + (-2)(-3) + (2)(6)$
$\mathbf{d} \cdot \mathbf{n} = 2 + 6 + 12 = 20$

3. Calculate the magnitudes of the direction vector and the normal vector.

$|\mathbf{d}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$
$|\mathbf{n}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$

4. Calculate the acute angle between the line and the plane.

Let $\theta$ be the acute angle between the line $l$ and the plane $p$.
$\sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}| |\mathbf{n}|}$
$\sin \theta = \frac{20}{(3)(7)} = \frac{20}{21}$
$\theta = \sin^{-1}\left(\frac{20}{21}\right) \approx 72.247^\circ$
$\theta \approx 72.2^\circ \text{ (to 1 d.p.)}$

1. Calculate the position vector of point $A$ from part (i) as it is needed to form the line equation.

$\text{Substitute } \mathbf{r} = \begin{pmatrix} 1 \\ 6 \\ -3 \end{pmatrix} + s\begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} \text{ into } p:$
$\left( \begin{pmatrix} 1+s \\ 6-2s \\ -3+2s \end{pmatrix} - \begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix} \right) \cdot \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix} = 0$
$\begin{pmatrix} s-2 \\ 6-2s \\ -3+2s \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix} = 0$
$2(s-2) - 3(6-2s) + 6(-3+2s) = 0$
$2s - 4 - 18 + 6s - 18 + 12s = 0 \implies 20s = 40 \implies s = 2$
$\vec{OA} = \begin{pmatrix} 1 \\ 6 \\ -3 \end{pmatrix} + 2\begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k}$

2. Determine the conditions for the direction vector of the new line.

$\text{Let the direction vector of the required line be } \mathbf{d}.$
$\text{Since the line lies in plane } p, \mathbf{d} \text{ is perpendicular to the normal vector of } p, \mathbf{n} = \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}.$
$\text{Since the line is perpendicular to } l, \mathbf{d} \text{ is perpendicular to the direction vector of } l, \mathbf{v} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}.$

3. Calculate the direction vector $\mathbf{d}$ using the vector cross product.

$\mathbf{d} = \mathbf{n} \times \mathbf{v} = \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix} \times \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$
$\mathbf{d} = \begin{pmatrix} (-3)(2) - (6)(-2) \\ (6)(1) - (2)(2) \\ (2)(-2) - (-3)(1) \end{pmatrix}$
$\mathbf{d} = \begin{pmatrix} -6 + 12 \\ 6 - 4 \\ -4 + 3 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \\ -1 \end{pmatrix}$

4. Formulate the final vector equation of the new line.

$\text{The line passes through } A(3, 2, 1) \text{ and has direction } \begin{pmatrix} 6 \\ 2 \\ -1 \end{pmatrix}.$
$\text{Vector equation: } \mathbf{r} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} + t \begin{pmatrix} 6 \\ 2 \\ -1 \end{pmatrix}, \text{ where } t \in \mathbb{R}$