平面 $\Pi$ 的方程为 $2x-3y+6z=7$，直线 $l$ 的方程为 $\mathbf{r}=8\mathbf{i}-2\mathbf{j}+8\mathbf{k}+t(3\mathbf{i}-\mathbf{j}+2\mathbf{k})$。 直线 $l$ 上的点 $A$ 和 $B$ 分别对应 $t=4$ 和 $t=2$。

1. Find the position vectors of points $A$ and $B$.

Substitute $t = 4$ and $t = 2$ into the equation of the line $l$ to find $\vec{OA}$ and $\vec{OB}$:
$\vec{OA} = \begin{pmatrix} 8 \\ -2 \\ 8 \end{pmatrix} + 4\begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 20 \\ -6 \\ 16 \end{pmatrix}$
$\vec{OB} = \begin{pmatrix} 8 \\ -2 \\ 8 \end{pmatrix} + 2\begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 14 \\ -4 \\ 12 \end{pmatrix}$

2. Determine the direction vectors $\vec{BA}$ and $\vec{BO}$.

$\vec{BA} = \vec{OA} - \vec{OB} = \begin{pmatrix} 20 \\ -6 \\ 16 \end{pmatrix} - \begin{pmatrix} 14 \\ -4 \\ 12 \end{pmatrix} = \begin{pmatrix} 6 \\ -2 \\ 4 \end{pmatrix}$
$\vec{BO} = -\vec{OB} = \begin{pmatrix} -14 \\ 4 \\ -12 \end{pmatrix}$

3. Use the scalar product to calculate $\angle ABO$.

$\cos \angle ABO = \frac{\vec{BA} \cdot \vec{BO}}{|\vec{BA}| |\vec{BO}|}$
$\vec{BA} \cdot \vec{BO} = (6)(-14) + (-2)(4) + (4)(-12) = -84 - 8 - 48 = -140$
$|\vec{BA}| = \sqrt{6^2 + (-2)^2 + 4^2} = \sqrt{56}$
$|\vec{BO}| = \sqrt{(-14)^2 + 4^2 + (-12)^2} = \sqrt{196 + 16 + 144} = \sqrt{356}$
$\cos \angle ABO = \frac{-140}{\sqrt{56}\sqrt{356}} = \frac{-140}{\sqrt{19936}}$
$\angle ABO = \cos^{-1}\left(\frac{-140}{\sqrt{19936}}\right) \approx 172.53^\circ \approx 172.5^\circ \text{ (1 d.p.)}$

1. Find the position vector of point $A$

Substitute $t=4$ into the equation of line $l$ to find the position vector of point $A$:
$\vec{OA} = \begin{pmatrix} 8 \\ -2 \\ 8 \end{pmatrix} + 4\begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 20 \\ -6 \\ 16 \end{pmatrix}$

2. Form the equation of the perpendicular line from $A$ to $\Pi$

The normal vector to the plane $\Pi$ is $\mathbf{n} = \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}$.
The line passing through $A$ and perpendicular to $\Pi$ has this normal vector as its direction vector. Its equation is:
$\mathbf{r} = \begin{pmatrix} 20 \\ -6 \\ 16 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}$

3. Find the point of intersection of the perpendicular line and the plane

Substitute the equation of the line into the cartesian equation of the plane $2x - 3y + 6z = 7$:
$2(20 + 2\mu) - 3(-6 - 3\mu) + 6(16 + 6\mu) = 7$
$40 + 4\mu + 18 + 9\mu + 96 + 36\mu = 7$
$154 + 49\mu = 7$
$49\mu = -147 \implies \mu = -3$
Substitute $\mu = -3$ back into the line equation to find the position vector of the foot of the perpendicular, $F$:
$\vec{OF} = \begin{pmatrix} 20 \\ -6 \\ 16 \end{pmatrix} - 3 \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix} = \begin{pmatrix} 14 \\ 3 \\ -2 \end{pmatrix}$
Thus, the coordinates of the foot of the perpendicular are $(14, 3, -2)$.

1. Identify the normal vectors to both planes.

From the Cartesian equation $2x - 3y + 6z = 7$, the normal vector to the plane $\Pi$ is $\mathbf{n}_1 = \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}$.
The equation $x = 0$ represents the $y$-$z$ plane, which has the normal vector $\mathbf{n}_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$.

2. Use the scalar product to find the acute angle between the normal vectors.

Let $\theta$ be the acute angle between the planes. This is equal to the acute angle between their normal vectors.
$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| |\mathbf{n}_2|}$
$\cos \theta = \frac{|(2)(1) + (-3)(0) + (6)(0)|}{\sqrt{2^2 + (-3)^2 + 6^2} \sqrt{1^2 + 0^2 + 0^2}}$
$\cos \theta = \frac{2}{\sqrt{4+9+36} (1)}$
$\cos \theta = \frac{2}{7}$
$\theta = \cos^{-1}\left(\frac{2}{7}\right) \approx 73.398^\circ$

1. Express the coordinates of the point $P$ in terms of $\lambda$.

Since $P$ lies on the line $l$ at $t=\lambda$, its position vector is:
$\vec{OP} = \begin{pmatrix} 8+3\lambda \\ -2-\lambda \\ 8+2\lambda \end{pmatrix}$

2. Apply the perpendicular distance formula from a point to a plane.

The equation of the plane $\Pi$ is $2x-3y+6z=7$.
The perpendicular distance $d$ from $P$ to $\Pi$ is given by:
$d = \frac{|2(8+3\lambda) - 3(-2-\lambda) + 6(8+2\lambda) - 7|}{\sqrt{2^2+(-3)^2+6^2}}$

3. Simplify the distance expression.

$d = \frac{|16 + 6\lambda + 6 + 3\lambda + 48 + 12\lambda - 7|}{\sqrt{49}}$
$d = \frac{|21\lambda + 63|}{7}$
$d = |3\lambda + 9| = 3|\lambda + 3|$

4. Set up and solve the modulus inequality.

We are given that the perpendicular distance is less than 3:
$3|\lambda + 3| < 3$
$|\lambda + 3| < 1$
$-1 < \lambda + 3 < 1$
$-4 < \lambda < -2$