The planes $\Pi_1$ and $\Pi_2$, which meet in the line $l$, have vector equations
$\mathbf{r} = 2\mathbf{i} + 4\mathbf{j} + 6\mathbf{k} + \theta_1(2\mathbf{i} + 3\mathbf{k}) + \phi_1(-4\mathbf{j} + 5\mathbf{k}),$
$\mathbf{r} = 2\mathbf{i} + 4\mathbf{j} + 6\mathbf{k} + \theta_2(3\mathbf{j} + \mathbf{k}) + \phi_2(-\mathbf{i} + \mathbf{j} + 2\mathbf{k}),$
respectively.

1. Identify a common point that lies on both planes.

By observing the parametric vector equations for $\Pi_1$ and $\Pi_2$, the position vector $\mathbf{a} = 2\mathbf{i} + 4\mathbf{j} + 6\mathbf{k}$ is present as the base point in both equations.
Thus, the point $(2, 4, 6)$ lies on both planes and must lie on the line of intersection $l$.

2. Find the normal vector to the first plane, $\Pi_1$.

The normal vector to $\Pi_1$, denoted as $\mathbf{n}_1$, is the cross product of its two direction vectors:
$\mathbf{n}_1 = \begin{pmatrix} 2 \\ 0 \\ 3 \end{pmatrix} \times \begin{pmatrix} 0 \\ -4 \\ 5 \end{pmatrix} = \begin{pmatrix} (0)(5) - (3)(-4) \\ (3)(0) - (2)(5) \\ (2)(-4) - (0)(0) \end{pmatrix} = \begin{pmatrix} 12 \\ -10 \\ -8 \end{pmatrix}$
We can use a simpler, parallel normal vector by factoring out $2$: $\mathbf{n}_1' = \begin{pmatrix} 6 \\ -5 \\ -4 \end{pmatrix}$.

3. Find the normal vector to the second plane, $\Pi_2$.

Similarly, the normal vector to $\Pi_2$, denoted as $\mathbf{n}_2$, is:
$\mathbf{n}_2 = \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix} \times \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} (3)(2) - (1)(1) \\ (1)(-1) - (0)(2) \\ (0)(1) - (3)(-1) \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \\ 3 \end{pmatrix}$

4. Find the direction vector of the line of intersection.

The line of intersection $l$ lies on both planes, so its direction vector $\mathbf{b}$ must be perpendicular to both normals.
$\mathbf{b} = \mathbf{n}_1' \times \mathbf{n}_2 = \begin{pmatrix} 6 \\ -5 \\ -4 \end{pmatrix} \times \begin{pmatrix} 5 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} (-5)(3) - (-4)(-1) \\ (-4)(5) - (6)(3) \\ (6)(-1) - (-5)(5) \end{pmatrix} = \begin{pmatrix} -19 \\ -38 \\ 19 \end{pmatrix}$
We can simplify this direction vector by dividing by $-19$ to get $\mathbf{b} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$.

5. Construct the vector equation of the line.

Using the common point $\mathbf{a}$ and the direction vector $\mathbf{b}$, the vector equation of the line $l$ is:
$\mathbf{r} = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} + t \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$
or expressed in unit vectors, $\mathbf{r} = 2\mathbf{i} + 4\mathbf{j} + 6\mathbf{k} + t(\mathbf{i} + 2\mathbf{j} - \mathbf{k})$, where $t \in \mathbb{R}$.

1. Compute the vector equation of line $l$ (required from the previous part).

From the given equations of $\Pi_1$ and $\Pi_2$, their normal vectors are:
$\mathbf{n}_1 = \begin{pmatrix} 2 \\ 0 \\ 3 \end{pmatrix} \times \begin{pmatrix} 0 \\ -4 \\ 5 \end{pmatrix} = \begin{pmatrix} 12 \\ -10 \\ -8 \end{pmatrix}$ and $\mathbf{n}_2 = \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix} \times \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \\ 3 \end{pmatrix}$.
The direction of line $l$ is parallel to $\mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} 12 \\ -10 \\ -8 \end{pmatrix} \times \begin{pmatrix} 5 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} -38 \\ -76 \\ 38 \end{pmatrix} = -38\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$.
By inspection of the given vector equations, both planes share the position vector $\begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}$.
Thus, the equation of $l$ is $\mathbf{r} = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} + t\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$.

2. Find a second direction vector parallel to $\Pi_3$.

Since plane $\Pi_3$ contains the line $l$, it contains the point $P(2, 4, 6)$ and has direction vector $\mathbf{d}_1 = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$.
It also passes through the given point $A(4, 3, 2)$. A second direction vector parallel to $\Pi_3$ is the displacement vector $\vec{PA}$:
$\mathbf{d}_2 = \vec{PA} = \begin{pmatrix} 4 \\ 3 \\ 2 \end{pmatrix} - \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ -4 \end{pmatrix}$

3. Formulate a vector equation of the plane $\Pi_3$.

Using point $A$ and the two direction vectors $\mathbf{d}_1$ and $\mathbf{d}_2$, the vector equation is:
$\mathbf{r} = \begin{pmatrix} 4 \\ 3 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ -1 \\ -4 \end{pmatrix}$
where $\lambda, \mu \in \mathbb{R}$.

4. Calculate the normal vector to $\Pi_3$.

The normal to $\Pi_3$ is given by the cross product $\mathbf{n}_3 = \mathbf{d}_1 \times \mathbf{d}_2$:
$\mathbf{n}_3 = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \times \begin{pmatrix} 2 \\ -1 \\ -4 \end{pmatrix} = \begin{pmatrix} (2)(-4) - (-1)(-1) \\ (-1)(2) - (1)(-4) \\ (1)(-1) - (2)(2) \end{pmatrix} = \begin{pmatrix} -9 \\ 2 \\ -5 \end{pmatrix}$
To simplify, we can take the negative of this vector as our normal vector: $\mathbf{n} = \begin{pmatrix} 9 \\ -2 \\ 5 \end{pmatrix}$.

5. Determine the Cartesian equation of $\Pi_3$.

Using the equation $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$, where $\mathbf{a}$ is the position vector of point $A$:
$\mathbf{r} \cdot \begin{pmatrix} 9 \\ -2 \\ 5 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 9 \\ -2 \\ 5 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 9 \\ -2 \\ 5 \end{pmatrix} = 4(9) + 3(-2) + 2(5) = 36 - 6 + 10 = 40$
The Cartesian equation of $\Pi_3$ is $9x - 2y + 5z = 40$.

1. Verify that the first two equations correspond to the Cartesian equations of planes $\Pi_1$ and $\Pi_2$.

For $\Pi_1$, a normal vector is $\mathbf{n}_1 = \begin{pmatrix} 2 \\ 0 \\ 3 \end{pmatrix} \times \begin{pmatrix} 0 \\ -4 \\ 5 \end{pmatrix} = \begin{pmatrix} 12 \\ -10 \\ -8 \end{pmatrix}$. Using a parallel normal vector $\begin{pmatrix} 6 \\ -5 \\ -4 \end{pmatrix}$, the Cartesian equation is $\mathbf{r} \cdot \begin{pmatrix} 6 \\ -5 \\ -4 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ -5 \\ -4 \end{pmatrix} = 12 - 20 - 24 = -32$, which gives $6x - 5y - 4z = -32$.

For $\Pi_2$, a normal vector is $\mathbf{n}_2 = \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix} \times \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \\ 3 \end{pmatrix}$. The Cartesian equation is $\mathbf{r} \cdot \begin{pmatrix} 5 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ -1 \\ 3 \end{pmatrix} = 10 - 4 + 18 = 24$, which gives $5x - y + 3z = 24$.

2. Identify the third equation as $\Pi_3$ and use the intersection of the planes to deduce the number of solutions.

The third equation, $9x - 2y + 5z = 40$, matches the Cartesian equation of $\Pi_3$ found in the previous part.

The given system of equations corresponds to finding the intersection of the three planes $\Pi_1$, $\Pi_2$, and $\Pi_3$.

We know that $\Pi_1$ and $\Pi_2$ intersect at the line $l$ (given), and $\Pi_3$ contains the line $l$ (from the previous part). Therefore, all three planes intersect at exactly the line $l$.

Since a line contains an infinite number of points, the system of equations has an infinite number of solutions.