平面 $\Pi_1$ 的方程为
$\mathbf{r} = \mathbf{i} + 2\mathbf{j} + \mathbf{k} + \theta(2\mathbf{j} - \mathbf{k}) + \phi(3\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}).$

1. Calculate the cross product of the plane's direction vectors to find a normal vector.

From the parametric equation of $\Pi_1$, the direction vectors are $\mathbf{d}_1 = \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix}$ and $\mathbf{d}_2 = \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}$.
Let $\mathbf{n}$ be the normal vector.
$\mathbf{n} = \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} \times \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}$
$= \begin{pmatrix} 2(-2) - (-1)(2) \\ (-1)(3) - (0)(-2) \\ 0(2) - 2(3) \end{pmatrix}$
$= \begin{pmatrix} -2 \\ -3 \\ -6 \end{pmatrix}$

2. Form the scalar product equation of the plane.

Using the position vector $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$ from the plane's equation and taking a parallel normal vector $\begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix}$ for simplicity:
$\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix} = 1(2) + 2(3) + 1(6) = 14$

3. Convert the scalar product equation to Cartesian form.

Substitute $\mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$:
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix} = 14$
$2x + 3y + 6z = 14 \quad \text{(shown)}$

1. Express the position vector of $P$ in terms of $\lambda$.

Since $P$ lies on the line $l$ where $t = \lambda$, its position vector is:
$\vec{OP} = \begin{pmatrix} 3 + 4\lambda \\ 8 + 6\lambda \\ 2 + 5\lambda \end{pmatrix}$

2. Apply the perpendicular distance formula from point $P$ to the plane $\Pi_1$.

From part [0], the Cartesian equation of $\Pi_1$ is $2x + 3y + 6z = 14$.
The perpendicular distance $D$ from $P$ to $\Pi_1$ is given by:
$D = \frac{|2(3 + 4\lambda) + 3(8 + 6\lambda) + 6(2 + 5\lambda) - 14|}{\sqrt{2^2 + 3^2 + 6^2}}$

3. Simplify the expression for the perpendicular distance.

$D = \frac{|6 + 8\lambda + 24 + 18\lambda + 12 + 30\lambda - 14|}{\sqrt{4 + 9 + 36}}$
$D = \frac{|42 + 56\lambda - 14|}{\sqrt{49}}$
$D = \frac{|28 + 56\lambda|}{7} = |4 + 8\lambda|$

4. Set up and solve the modulus inequality to find the range of $\lambda$.

The perpendicular distance is not greater than 4, which means $D \le 4$:
$|4 + 8\lambda| \le 4$
$-4 \le 4 + 8\lambda \le 4$
$-8 \le 8\lambda \le 0$
$-1 \le \lambda \le 0$

1. State the normal vector to the first plane, $\Pi_1$.

From part [0], the Cartesian equation of the plane is $2x + 3y + 6z = 14$.
The normal vector to $\Pi_1$ can be read directly from the coefficients:
$\mathbf{n}_1 = \begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix}$

2. Identify two non-parallel direction vectors that are parallel to $\Pi_2$.

Plane $\Pi_2$ contains line $l$, so one direction vector is the direction of $l$:
$\mathbf{d} = \begin{pmatrix} 4 \\ 6 \\ 5 \end{pmatrix}$
$\Pi_2$ also contains a specific point $A(1, 2, 1)$ and the position vector $B(3, 8, 2)$ which lies on $l$. The vector connecting these two points is also parallel to $\Pi_2$:
$\vec{AB} = \begin{pmatrix} 3 \\ 8 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 6 \\ 1 \end{pmatrix}$

3. Determine a normal vector to $\Pi_2$ using the cross product.

The normal vector to $\Pi_2$ is perpendicular to both $\mathbf{d}$ and $\vec{AB}$:
$\mathbf{n}_2 = \mathbf{d} \times \vec{AB} = \begin{pmatrix} 4 \\ 6 \\ 5 \end{pmatrix} \times \begin{pmatrix} 2 \\ 6 \\ 1 \end{pmatrix} = \begin{pmatrix} (6)(1) - (5)(6) \\ (5)(2) - (4)(1) \\ (4)(6) - (6)(2) \end{pmatrix} = \begin{pmatrix} -24 \\ 6 \\ 12 \end{pmatrix}$
Dividing by a common factor of 6 to simplify, we can use a parallel normal vector:
$\mathbf{n}_2' = \begin{pmatrix} -4 \\ 1 \\ 2 \end{pmatrix}$

4. Calculate the acute angle between $\Pi_1$ and $\Pi_2$.

Let $\theta$ be the acute angle between the planes, which is the acute angle between their normal vectors.
$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2'|}{|\mathbf{n}_1| |\mathbf{n}_2'|}$
$\cos \theta = \frac{\left| \begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix} \cdot \begin{pmatrix} -4 \\ 1 \\ 2 \end{pmatrix} \right|}{\sqrt{2^2 + 3^2 + 6^2} \sqrt{(-4)^2 + 1^2 + 2^2}}$
$\cos \theta = \frac{|-8 + 3 + 12|}{\sqrt{49} \sqrt{21}} = \frac{7}{7\sqrt{21}} = \frac{1}{\sqrt{21}}$
$\theta = \cos^{-1}\left(\frac{1}{\sqrt{21}}\right) \approx 77.4^\circ \text{ (to 1 d.p.)}$