点 A 的坐标为 (1, -2, 4)，平面 $\pi_1$ 的方程为 $2x - y + 2z = 5.$

1. Substitute the coordinates of point $A$ and the plane equation into the perpendicular distance formula.

The shortest distance $D$ from a point $(x_1, y_1, z_1)$ to a plane $ax + by + cz = d$ is given by $D = \frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}}$.

$D = \frac{|2(1) - (-2) + 2(4) - 5|}{\sqrt{2^2 + (-1)^2 + 2^2}}$

2. Evaluate the expression to find the exact distance.

$D = \frac{|2 + 2 + 8 - 5|}{\sqrt{9}}$
$D = \frac{7}{3}$

1. Identify the normal vectors of the two planes to establish how to find the direction vector of the line.

The normal vectors of planes $\pi_1$ and $\pi_2$ are $\mathbf{n}_1 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ and $\mathbf{n}_2 = \begin{pmatrix} 1 \\ 3 \\ -a \end{pmatrix}$.

The line of intersection $l$ lies on both planes, so its direction vector $\mathbf{d}$ must be perpendicular to both normal vectors. Therefore, $\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2$.

2. Calculate the cross product to find the direction vector of the line.

$\mathbf{d} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 3 \\ -a \end{pmatrix} = \begin{pmatrix} (-1)(-a) - (2)(3) \\ (2)(1) - (2)(-a) \\ (2)(3) - (-1)(1) \end{pmatrix} = \begin{pmatrix} a - 6 \\ 2a + 2 \\ 7 \end{pmatrix}$

3. Find the coordinates of a common point on both planes by setting one variable to zero.

Let $z = 0$. The Cartesian equations of the planes become a linear system:
$2x - y = 5 \quad \text{--- (1)}$
$x + 3y = 3 \quad \text{--- (2)}$

From (1), $y = 2x - 5$. Substituting this into (2):
$x + 3(2x - 5) = 3$
$7x - 15 = 3 \implies 7x = 18 \implies x = \frac{18}{7}$

Then, $y = 2\left(\frac{18}{7}\right) - 5 = \frac{36}{7} - \frac{35}{7} = \frac{1}{7}$.

A point on the line $l$ is $\left( \frac{18}{7}, \frac{1}{7}, 0 \right)$.

4. Construct the final vector equation of the line.

The vector equation of line $l$ is:
$\mathbf{r} = \begin{pmatrix} 18/7 \\ 1/7 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} a - 6 \\ 2a + 2 \\ 7 \end{pmatrix}, \text{ where } \lambda \in \mathbb{R}$

1. Determine the direction vector of line $l$ by taking the cross product of the normal vectors of $\pi_1$ and $\pi_2$ (as required from part b).

The normal vectors of $\pi_1$ and $\pi_2$ are $\mathbf{n}_1 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ and $\mathbf{n}_2 = \begin{pmatrix} 1 \\ 3 \\ -a \end{pmatrix}$.
The direction vector of the line of intersection, $l$, is given by $\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2$.
$\mathbf{d} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 3 \\ -a \end{pmatrix} = \begin{pmatrix} (-1)(-a) - (2)(3) \\ (2)(1) - (2)(-a) \\ (2)(3) - (-1)(1) \end{pmatrix} = \begin{pmatrix} a - 6 \\ 2a + 2 \\ 7 \end{pmatrix}$

2. Set up the perpendicularity condition between the line's direction vector and the normal vector of $\pi_3$.

The normal vector of $\pi_3$ is $\mathbf{n}_3 = \begin{pmatrix} b \\ -2 \\ 4 \end{pmatrix}$.
For the line $l$ to be parallel to the plane $\pi_3$, its direction vector $\mathbf{d}$ must be perpendicular to the plane's normal vector $\mathbf{n}_3$. Thus, their dot product must be zero: $\mathbf{d} \cdot \mathbf{n}_3 = 0$.
$\begin{pmatrix} a - 6 \\ 2a + 2 \\ 7 \end{pmatrix} \cdot \begin{pmatrix} b \\ -2 \\ 4 \end{pmatrix} = 0$

3. Expand the dot product and factorise to obtain the required equation.

Evaluating the scalar product:
$b(a - 6) - 2(2a + 2) + 4(7) = 0$
$ab - 6b - 4a - 4 + 28 = 0$
$ab - 6b - 4a + 24 = 0$
By grouping the terms:
$b(a - 6) - 4(a - 6) = 0$
$(a - 6)(b - 4) = 0 \quad \text{(shown)}$

1. Deduce the value of $a$ using the necessary condition for a triangular prism and checking for parallel planes.

For the three planes to form a triangular prism, they must intersect in pairs forming three parallel lines. This requires the line of intersection $l$ of $\pi_1$ and $\pi_2$ to be parallel to $\pi_3$. From part (c), this condition means $(a-6)(b-4) = 0$.

We must check if $b = 4$:
If $b = 4$, the normal vector of $\pi_3$ is $\mathbf{n}_3 = \begin{pmatrix} 4 \\ -2 \\ 4 \end{pmatrix} = 2 \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} = 2\mathbf{n}_1$.
This would mean $\pi_1$ and $\pi_3$ are parallel. Since the question requires all planes to be non-parallel, $b \neq 4$.
Therefore, to satisfy $(a-6)(b-4) = 0$, we must have $a = 6$.

2. Find the condition on $b$ such that $\pi_2$ and $\pi_3$ are not parallel.

With $a = 6$, the normal vector of $\pi_2$ is $\mathbf{n}_2 = \begin{pmatrix} 1 \\ 3 \\ -6 \end{pmatrix}$.
(Note that $\mathbf{n}_1$ and $\mathbf{n}_2$ are non-parallel since their components are not proportional).

For $\pi_2$ and $\pi_3$ to be parallel, $\mathbf{n}_3 = k\mathbf{n}_2$ for some constant $k$:
$\begin{pmatrix} b \\ -2 \\ 4 \end{pmatrix} = k \begin{pmatrix} 1 \\ 3 \\ -6 \end{pmatrix}$
From the $y$-components, $-2 = 3k \implies k = -\frac{2}{3}$.
Checking the $z$-components: $-6\left(-\frac{2}{3}\right) = 4$, which is consistent.
Thus, $b = k(1) = -\frac{2}{3}$.
Since all planes must be non-parallel, we require $b \neq -\frac{2}{3}$.

3. Find equations representing the line of intersection $l$ between $\pi_1$ and $\pi_2$.

Substitute $a=6$ into $\pi_2$ and solve the system for $\pi_1$ and $\pi_2$ to find any point on their line of intersection $l$:
$2x - y + 2z = 5 \quad \text{--- (1)}$
$x + 3y - 6z = 3 \quad \text{--- (2)}$
To eliminate $y$ and $z$, we can compute $3 \times (1) + (2)$:
$3(2x - y + 2z) + (x + 3y - 6z) = 3(5) + 3$
$7x = 18 \implies x = \frac{18}{7}$
Substitute $x = \frac{18}{7}$ into (2):
$\frac{18}{7} + 3y - 6z = 3$
$3(y - 2z) = 3 - \frac{18}{7} = \frac{3}{7} \implies y - 2z = \frac{1}{7}$
The line $l$ is defined by the equations $x = \frac{18}{7}$ and $y - 2z = \frac{1}{7}$.

4. Determine the final condition on $b$ so that the planes do not have any point in common.

For the three planes to not have a common point, the line $l$ must not intersect or lie on $\pi_3$. Since $l \parallel \pi_3$ is already established, we just need to ensure $l$ does not lie entirely on $\pi_3$.
Substitute the constraint equations of $l$ into $\pi_3$:
$bx - 2y + 4z = 3 \implies b\left(\frac{18}{7}\right) - 2(y - 2z) = 3$
Substitute $y - 2z = \frac{1}{7}$:
$\frac{18b}{7} - 2\left(\frac{1}{7}\right) = 3$
$18b - 2 = 21 \implies 18b = 23 \implies b = \frac{23}{18}$
If $b = \frac{23}{18}$, the three planes intersect at a common line. Thus, we must have $b \neq \frac{23}{18}$.

Conclusion: The conditions are $a = 6$, and $b \neq 4$, $b \neq -\frac{2}{3}$, $b \neq \frac{23}{18}$.