With reference to the origin O, the point A has position vector $-\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}$. The plane $\pi$ and the line $l$ have equations $\mathbf{r} = \begin{pmatrix}4 \\ 2 \\ 1\end{pmatrix} + \lambda \begin{pmatrix}-1 \\ 3 \\ 2\end{pmatrix} + \mu \begin{pmatrix}2 \\ 3 \\ -4\end{pmatrix}$ and $\mathbf{r} = \begin{pmatrix}2 \\ 1 \\ 1\end{pmatrix} + \beta \begin{pmatrix}3 \\ t \\ -2\end{pmatrix},$ respectively, where $\lambda$, $\mu$ and $\beta$ are parameters and $t$ is a real value.

1. Find the normal vector and Cartesian equation of the plane $\pi$.

The normal vector to the plane $\pi$ is the cross product of its direction vectors:
$\mathbf{n} = \begin{pmatrix}-1 \\ 3 \\ 2\end{pmatrix} \times \begin{pmatrix}2 \\ 3 \\ -4\end{pmatrix} = \begin{pmatrix} (3)(-4) - (2)(3) \\ (2)(2) - (-1)(-4) \\ (-1)(3) - (3)(2) \end{pmatrix} = \begin{pmatrix} -18 \\ 0 \\ -9 \end{pmatrix}$
Taking a simpler parallel normal vector $\mathbf{n}' = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$, the Cartesian equation of the plane $\pi$ is:
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = 8 + 0 + 1 = 9$

2. Substitute the equation of line $l$ into the plane equation to find the intersection condition.

Substitute the equation of $l$, $\mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} + \beta \begin{pmatrix} 3 \\ t \\ -2 \end{pmatrix}$, into the plane equation:
$\left[ \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} + \beta \begin{pmatrix} 3 \\ t \\ -2 \end{pmatrix} \right] \cdot \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = 9$
$(4 + 0 + 1) + \beta (6 + 0 - 2) = 9$
$5 + 4\beta = 9$
Since $t$ is eliminated and we obtain $4\beta = 4$ (which yields a unique solution $\beta = 1$), a unique point of intersection exists regardless of the value of $t$. Thus, $t$ can be any real number.

1. Find the scalar product equation of the plane $\pi$ by calculating its normal vector.

The normal vector to the plane $\pi$ is the cross product of its two direction vectors:
$\mathbf{n} = \begin{pmatrix}-1 \\ 3 \\ 2\end{pmatrix} \times \begin{pmatrix}2 \\ 3 \\ -4\end{pmatrix} = \begin{pmatrix} (3)(-4) - (2)(3) \\ (2)(2) - (-1)(-4) \\ (-1)(3) - (3)(2) \end{pmatrix} = \begin{pmatrix} -18 \\ 0 \\ -9 \end{pmatrix} = -9\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$
Taking the normal vector as $\mathbf{n} = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$, the scalar product equation of $\pi$ is:
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = 8 + 0 + 1 = 9$

2. Substitute the equation of line $l$ into the plane equation to find the intersection point $B$.

Substitute the vector equation of line $l$ into the scalar product equation of $\pi$:
$\left[ \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} + \beta \begin{pmatrix} 3 \\ t \\ -2 \end{pmatrix} \right] \cdot \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = 9$
$(2 + 3\beta)(2) + (1 + t\beta)(0) + (1 - 2\beta)(1) = 9$
$4 + 6\beta + 1 - 2\beta = 9$
$5 + 4\beta = 9$
$4\beta = 4 \implies \beta = 1$

Substitute $\beta = 1$ back into the equation of line $l$ to find the position vector of $B$:
$\vec{OB} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} + (1)\begin{pmatrix} 3 \\ t \\ -2 \end{pmatrix} = \begin{pmatrix} 5 \\ 1+t \\ -1 \end{pmatrix}$
The coordinates of $B$ are $(5, 1+t, -1)$.

1. Equate the position vector of point $A$ to the equation of line $l$.

$\begin{pmatrix}-1 \\ -3 \\ 3\end{pmatrix} = \begin{pmatrix}2 \\ 1 \\ 1\end{pmatrix} + \beta \begin{pmatrix}3 \\ t \\ -2\end{pmatrix}$

2. Solve for $\beta$ using the $x$-component and substitute into the $y$-component to find $t$.

From the $x$-components:
$-1 = 2 + 3\beta \implies 3\beta = -3 \implies \beta = -1$

(Checking with $z$-components: $3 = 1 - 2(-1) = 3$, consistent.)

From the $y$-components:
$-3 = 1 + \beta t$
Substitute $\beta = -1$:
$-3 = 1 - t \implies t = 4 \text{ (shown)}$

1. Find a normal vector to the plane $\pi$.

The normal vector $\mathbf{n}$ can be found using the cross product of the two direction vectors of the plane $\pi$:

$\mathbf{n} = \begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix} \times \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix}$

$\mathbf{n} = \begin{pmatrix} (3)(-4) - (2)(3) \\ (2)(2) - (-1)(-4) \\ (-1)(3) - (3)(2) \end{pmatrix} = \begin{pmatrix} -18 \\ 0 \\ -9 \end{pmatrix}$

Taking a simpler parallel vector by dividing by $-9$, we use $\mathbf{n} = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$.

2. Formulate the scalar product equation of the plane and the vector equation of the normal line through $A$.

The equation of plane $\pi$ is $\mathbf{r} \cdot \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = 8 + 0 + 1 = 9$.

The normal line passing through $A$ perpendicular to $\pi$ has the equation:
$l_A: \mathbf{r} = \begin{pmatrix} -1 \\ -3 \\ 3 \end{pmatrix} + \alpha \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}, \text{ where } \alpha \in \mathbb{R}$

3. Substitute the line equation into the plane equation to find the foot of the perpendicular, $F$.

Substitute $l_A$ into the equation of $\pi$:

$\left[ \begin{pmatrix} -1 \\ -3 \\ 3 \end{pmatrix} + \alpha \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} \right] \cdot \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = 9$

$(-2 + 0 + 3) + \alpha(2^2 + 0^2 + 1^2) = 9$

$1 + 5\alpha = 9 \implies 5\alpha = 8 \implies \alpha = \frac{8}{5}$

Substitute $\alpha = \frac{8}{5}$ back into the line equation to find the position vector of $F$:

$\overrightarrow{OF} = \begin{pmatrix} -1 \\ -3 \\ 3 \end{pmatrix} + \frac{8}{5} \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 + 16/5 \\ -3 + 0 \\ 3 + 8/5 \end{pmatrix} = \begin{pmatrix} 11/5 \\ -3 \\ 23/5 \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 11 \\ -15 \\ 23 \end{pmatrix}$

1. Concisely recover the coordinates of intersection point $B$ and foot of perpendicular $F$ from earlier parts.

The normal to $\pi$ is $\mathbf{n} = \begin{pmatrix}-1\\3\\2\end{pmatrix} \times \begin{pmatrix}2\\3\\-4\end{pmatrix} = \begin{pmatrix}-18\\0\\-9\end{pmatrix} \equiv \begin{pmatrix}2\\0\\1\end{pmatrix}$, so $\pi$ has equation $\mathbf{r} \cdot \begin{pmatrix}2\\0\\1\end{pmatrix} = 9$.
Substituting $l$ (with $t=4$) into $\pi$: $\begin{pmatrix}2+3\beta\\1+4\beta\\1-2\beta\end{pmatrix} \cdot \begin{pmatrix}2\\0\\1\end{pmatrix} = 9 \implies 4 + 6\beta + 1 - 2\beta = 9 \implies \beta = 1$.
Thus, $B$ has position vector $\vec{OB} = \begin{pmatrix}5\\5\\-1\end{pmatrix}$.
The line $AF$ passes through $A(-1, -3, 3)$ and is parallel to $\mathbf{n}$, so $\mathbf{r} = \begin{pmatrix}-1\\-3\\3\end{pmatrix} + \alpha \begin{pmatrix}2\\0\\1\end{pmatrix}$.
Substituting into $\pi$: $2(-1+2\alpha) + (3+\alpha) = 9 \implies 5\alpha = 8 \implies \alpha = \frac{8}{5}$.
Thus, $F$ has position vector $\vec{OF} = \begin{pmatrix}11/5\\-3\\23/5\end{pmatrix}$.

2. Use the geometric properties of the plane $ABF$ to determine the point $B'$ where line $m$ intersects $\pi$.

Since line $m$ lies on the plane containing $A$, $B$ and $F$, and makes the same angle with $\pi$ as $l$, $m$ must intersect the line $BF$ (the projection of $l$ on $\pi$) at a point $B'$.
Because $AF \perp \pi$, $\triangle ABB'$ forms an isosceles triangle with altitude $AF$. Hence, $F$ is the midpoint of $BB'$.
$\vec{FB'} = \vec{BF} \implies \vec{OB'} = 2\vec{OF} - \vec{OB}$
$\vec{OB'} = 2 \begin{pmatrix} 11/5 \\ -3 \\ 23/5 \end{pmatrix} - \begin{pmatrix} 5 \\ 5 \\ -1 \end{pmatrix} = \begin{pmatrix} 22/5 \\ -6 \\ 46/5 \end{pmatrix} - \begin{pmatrix} 25/5 \\ 5 \\ -5/5 \end{pmatrix} = \begin{pmatrix} -3/5 \\ -11 \\ 51/5 \end{pmatrix}$

3. Find the direction vector of $m$ and deduce its cartesian equation.

The direction vector of $m$ is given by $\vec{AB'}$:
$\vec{AB'} = \vec{OB'} - \vec{OA} = \begin{pmatrix} -3/5 \\ -11 \\ 51/5 \end{pmatrix} - \begin{pmatrix} -1 \\ -3 \\ 3 \end{pmatrix} = \begin{pmatrix} 2/5 \\ -8 \\ 36/5 \end{pmatrix}$
To simplify the direction vector, we can factor out a scalar multiplier. Multiplying by $\frac{5}{2}$ gives an integer direction vector:
$\mathbf{d}_m = \begin{pmatrix} 1 \\ -20 \\ 18 \end{pmatrix}$
Using point $A(-1, -3, 3)$ and direction vector $\mathbf{d}_m$, the cartesian equation for $m$ is:
$\frac{x + 1}{1} = \frac{y + 3}{-20} = \frac{z - 3}{18}$