The point A has coordinates (1, -2, 4) and the plane $\pi_{1}$ has equation $2x - y + 2z = 5$. The plane $\pi_{2}$ has equation $x+3y-az=3$, where $a$ is a constant. The plane $\pi_{3}$ has equation $bx-2y+4z=3$, where $b$ is a constant.

1. Set up the formula for the perpendicular distance from a point to a plane.

Using the distance formula $d = \frac{|\mathbf{a} \cdot \mathbf{n} - D|}{|\mathbf{n}|}$, where $\mathbf{a} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix}$, $\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$, and $D = 5$:
$\text{Distance} = \frac{\left| \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} - 5 \right|}{\sqrt{2^2 + (-1)^2 + 2^2}}$

2. Evaluate the scalar product and magnitude to find the exact shortest distance.

$\text{Distance} = \frac{| (1)(2) + (-2)(-1) + (4)(2) - 5 |}{\sqrt{4 + 1 + 4}}$
$= \frac{| 2 + 2 + 8 - 5 |}{\sqrt{9}}$
$= \frac{|12 - 5|}{3} = \frac{7}{3}$

1. Find the direction vector of the line of intersection

The normal vectors of the planes 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 $\mathbf{d}$ of the line of intersection is perpendicular to both normal vectors, so we compute their cross product:
$\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 3 \\ -a \end{pmatrix}$
$\mathbf{d} = \begin{pmatrix} (-1)(-a) - (2)(3) \\ (2)(1) - (2)(-a) \\ (2)(3) - (-1)(1) \end{pmatrix} = \begin{pmatrix} a - 6 \\ 2a + 2 \\ 7 \end{pmatrix}$

2. Find a position vector of a point on the line

We find a common point on both planes by setting one of the coordinates to a constant. Setting $z = 0$ provides a point independent of $a$:
$2x - y = 5 \quad \text{--- (1)}$
$x + 3y = 3 \quad \text{--- (2)}$
From (2), $x = 3 - 3y$. Substituting this into (1):
$2(3 - 3y) - y = 5$
$6 - 7y = 5 \implies 7y = 1 \implies y = \frac{1}{7}$
Substituting $y = \frac{1}{7}$ back to find $x$:
$x = 3 - 3\left(\frac{1}{7}\right) = \frac{18}{7}$
So, a point on the line is $\left( \frac{18}{7}, \frac{1}{7}, 0 \right)$.

3. State the vector equation of the line

Using the position vector of the point and the direction vector, the vector equation of the line $l$ is:
$\mathbf{r} = \begin{pmatrix} \frac{18}{7} \\ \frac{1}{7} \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} a - 6 \\ 2a + 2 \\ 7 \end{pmatrix}, \text{ where } \lambda \in \mathbb{R}$

1. Find the direction vector of line $l$ using the normal vectors of $\pi_1$ and $\pi_2$.

The line $l$ lies on both $\pi_1$ and $\pi_2$, so its direction vector $\mathbf{d}$ is perpendicular to both normal vectors $\mathbf{n}_1$ and $\mathbf{n}_2$.

2. Apply the condition for the line to be parallel to $\pi_3$ and manipulate to show the given result.

Since line $l$ is parallel to plane $\pi_3$, its direction vector $\mathbf{d}$ must be perpendicular to the normal vector of $\pi_3$, which is $\mathbf{n}_3 = \begin{pmatrix} b \\ -2 \\ 4 \end{pmatrix}$.
Therefore, $\mathbf{d} \cdot \mathbf{n}_3 = 0$.

Factorising by grouping:

1. Deduce the required value of $a$ using the prism condition and the non-parallel planes condition.

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

The normal vectors are $\mathbf{n}_1 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ and $\mathbf{n}_3 = \begin{pmatrix} b \\ -2 \\ 4 \end{pmatrix}$.
If $b = 4$, $\mathbf{n}_3 = 2\mathbf{n}_1$, making $\pi_1$ and $\pi_3$ parallel.
Since the question specifies that all planes must be non-parallel, we must restrict $b \neq 4$.
Therefore, to satisfy $(a-6)(b-4) = 0$, we must have $a = 6$.

2. Find the restriction on $b$ that ensures $\pi_2$ and $\pi_3$ are not parallel.

With $a = 6$, the normal vector of $\pi_2$ is $\mathbf{n}_2 = \begin{pmatrix} 1 \\ 3 \\ -6 \end{pmatrix}$.
For $\pi_2$ and $\pi_3$ to be parallel, their normals must be scalar multiples: $\mathbf{n}_2 = k\mathbf{n}_3 \implies \begin{pmatrix} 1 \\ 3 \\ -6 \end{pmatrix} = k\begin{pmatrix} b \\ -2 \\ 4 \end{pmatrix}$.
Comparing the $y$-components: $3 = -2k \implies k = -\frac{3}{2}$.
Substituting $k$ into the $x$-component: $1 = kb \implies b = -\frac{2}{3}$.
Since all planes must be non-parallel, we must restrict $b \neq -\frac{2}{3}$.

3. Formulate an intersection equation to ensure the planes do not intersect at a common line.

For the planes to have no point in common, the intersection line $l$ must not lie on $\pi_3$.
Using $a=6$, find a point on line $l$ by substituting $z=0$ into $\pi_1$ and $\pi_2$:
$2x - y = 5$
$x + 3y = 3$
Solving gives $x = \frac{18}{7}, y = \frac{1}{7}$. So $\left(\frac{18}{7}, \frac{1}{7}, 0\right)$ is a point on $l$.
The direction of $l$ is $\mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 3 \\ -6 \end{pmatrix} = \begin{pmatrix} 0 \\ 14 \\ 7 \end{pmatrix} \parallel \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}$.
Thus, $l$ has the vector equation $\mathbf{r} = \begin{pmatrix} 18/7 \\ 1/7 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}$.
Substitute this into the equation of $\pi_3$ ($bx - 2y + 4z = 3$):
$b\left(\frac{18}{7}\right) - 2\left(\frac{1}{7} + 2\lambda\right) + 4(\lambda) = 3$

4. Solve for $b$ to find the final restriction preventing infinitely many common points.

Simplifying the intersection equation:
$\frac{18b}{7} - \frac{2}{7} - 4\lambda + 4\lambda = 3$
$\frac{18b - 2}{7} = 3 \implies 18b - 2 = 21 \implies 18b = 23 \implies b = \frac{23}{18}$
If $b = \frac{23}{18}$, the equation is satisfied for all $\lambda$, meaning line $l$ lies entirely on $\pi_3$ and the planes would intersect at a common line.
To ensure there are no common points at all, $b \neq \frac{23}{18}$.
Combining all conditions: $a = 6$ and $b \neq 4, -\frac{2}{3}, \frac{23}{18}$.