The plane $\Pi$ has equation $2x-3y+6z=7$ and the line $l$ has equation $\mathbf{r}=8\mathbf{i}-2\mathbf{j}+8\mathbf{k}+t(3\mathbf{i}-\mathbf{j}+2\mathbf{k}).\n\nThe points A and B on$l$are given by$t=4$and$t=2$ respectively.

1. Find the position vector of $B$ and the vector $\vec{BO}$.

Given $B$ lies on $l$ at $t = 2$,
$\vec{OB} = \begin{pmatrix} 8 \\ -2 \\ 8 \end{pmatrix} + 2\begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 14 \\ -4 \\ 12 \end{pmatrix}$
Since the angle is at vertex $B$, we need the vector originating from $B$ towards $O$:
$\vec{BO} = -\vec{OB} = \begin{pmatrix} -14 \\ 4 \\ -12 \end{pmatrix}$

2. Find the vector $\vec{BA}$.

Given $A$ lies on $l$ at $t = 4$,
$\vec{OA} = \begin{pmatrix} 8 \\ -2 \\ 8 \end{pmatrix} + 4\begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 20 \\ -6 \\ 16 \end{pmatrix}$
$\vec{BA} = \vec{OA} - \vec{OB} = \begin{pmatrix} 20 \\ -6 \\ 16 \end{pmatrix} - \begin{pmatrix} 14 \\ -4 \\ 12 \end{pmatrix} = \begin{pmatrix} 6 \\ -2 \\ 4 \end{pmatrix}$
(Note: $\vec{BA}$ is in the same direction as the line's direction vector $\mathbf{d} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}$ since $4 > 2$, and $2\mathbf{d}$ could equivalently be used).

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

The angle $ABO$ is the angle between vectors $\vec{BA}$ and $\vec{BO}$:
$\cos(\angle ABO) = \frac{\vec{BA} \cdot \vec{BO}}{|\vec{BA}||\vec{BO}|}$
$\cos(\angle ABO) = \frac{\begin{pmatrix} 6 \\ -2 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} -14 \\ 4 \\ -12 \end{pmatrix}}{\sqrt{6^2 + (-2)^2 + 4^2}\sqrt{(-14)^2 + 4^2 + (-12)^2}}$
$\cos(\angle ABO) = \frac{6(-14) + (-2)(4) + 4(-12)}{\sqrt{36 + 4 + 16}\sqrt{196 + 16 + 144}}$
$\cos(\angle ABO) = \frac{-140}{\sqrt{56}\sqrt{356}} = \frac{-140}{\sqrt{19936}}$
$\angle ABO = \cos^{-1}\left(\frac{-140}{\sqrt{19936}}\right) \approx 172.5^\circ$

1. Find the position vector of point $A$.

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

2. Formulate the equation of the line passing through $A$ and perpendicular to $\Pi$.

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

3. Substitute the line equation into the plane equation to find the parameter for the foot of perpendicular.

Since the foot of perpendicular $F$ lies on $\Pi$, its position vector satisfies the plane equation:
$\begin{pmatrix} 20+2\lambda \\ -6-3\lambda \\ 16+6\lambda \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix} = 7$
$2(20 + 2\lambda) - 3(-6 - 3\lambda) + 6(16 + 6\lambda) = 7$
$40 + 4\lambda + 18 + 9\lambda + 96 + 36\lambda = 7$
$49\lambda + 154 = 7$
$49\lambda = -147 \implies \lambda = -3$

4. Find the coordinates of the foot of perpendicular.

Substitute $\lambda = -3$ back into the line equation:
$\vec{OF} = \begin{pmatrix} 20 \\ -6 \\ 16 \end{pmatrix} - 3 \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix} = \begin{pmatrix} 14 \\ 3 \\ -2 \end{pmatrix}$
The coordinates of the foot of perpendicular are $(14, 3, -2)$.

1. Identify the normal vectors of both planes.

The normal vector of $\Pi$ is $\mathbf{n}_1 = \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}$.
The plane $x=0$ (the $y$-$z$ plane) has normal vector $\mathbf{n}_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$.

2. Calculate the acute angle between the planes using the scalar product of their normal vectors.

Let $\theta$ be the acute angle between the planes.
$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| |\mathbf{n}_2|}$
$\cos \theta = \frac{|(2)(1) + (-3)(0) + (6)(0)|}{\sqrt{2^2 + (-3)^2 + 6^2} \sqrt{1^2 + 0^2 + 0^2}}$
$\cos \theta = \frac{2}{\sqrt{49} (1)} = \frac{2}{7}$
$\theta = \cos^{-1}\left(\frac{2}{7}\right) \approx 73.398^\circ$
$\theta \approx 73.4^\circ \text{ (to 1 d.p.)}$

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

Substitute $t = \lambda$ into the equation of line $l$:
$\vec{OP} = \begin{pmatrix} 8 \\ -2 \\ 8 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 + 3\lambda \\ -2 - \lambda \\ 8 + 2\lambda \end{pmatrix}$

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

The normal vector to the plane $\Pi$ is $\mathbf{n} = \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}$.
Distance from $P$ to $\Pi = \frac{|\vec{OP} \cdot \mathbf{n} - D|}{|\mathbf{n}|}$
$= \frac{\left| \begin{pmatrix} 8 + 3\lambda \\ -2 - \lambda \\ 8 + 2\lambda \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix} - 7 \right|}{\sqrt{2^2 + (-3)^2 + 6^2}}$

3. Simplify the expression for the perpendicular distance.

$= \frac{|2(8 + 3\lambda) - 3(-2 - \lambda) + 6(8 + 2\lambda) - 7|}{\sqrt{4 + 9 + 36}}$
$= \frac{|16 + 6\lambda + 6 + 3\lambda + 48 + 12\lambda - 7|}{\sqrt{49}}$
$= \frac{|63 + 21\lambda|}{7} = |9 + 3\lambda|$

4. Formulate and solve the modulus inequality.

Given that the distance is less than 3:
$|9 + 3\lambda| < 3$
$-3 < 9 + 3\lambda < 3$
$-12 < 3\lambda < -6$
$-4 < \lambda < -2$