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

1. Identify the position vector of $A$ and the normal vector of the plane, and apply the shortest distance formula.

Let the position vector of $A$ be $\mathbf{a} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix}$ and the normal vector to the plane $\pi_1$ be $\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$.
The shortest distance $D$ from a point to a plane $\mathbf{r} \cdot \mathbf{n} = p$ is given by $D = \frac{|\mathbf{a} \cdot \mathbf{n} - p|}{|\mathbf{n}|}$.
$D = \frac{\left| \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} - 5 \right|}{\sqrt{2^2 + (-1)^2 + 2^2}}$

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

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

1. Find the direction vector of the line of intersection by taking the cross product of the normal vectors of the two planes.

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 line of intersection $l$ lies on both planes, so its direction vector $\mathbf{d}$ is perpendicular to both normal vectors, given by $\mathbf{n}_1 \times \mathbf{n}_2$.

2. Evaluate the cross product to obtain the direction vector in terms of $a$.

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

3. Find a specific point on the line $l$ by substituting a value for one of the variables.

To find a point on $l$, we can find a common solution to the equations of $\pi_1$ and $\pi_2$. To avoid dealing with $a$ in the coordinates, we can set $z = 0$:
$\begin{aligned} 2x - y &= 5 \quad \text{--- (1)} \\ x + 3y &= 3 \quad \text{--- (2)} \end{aligned}$

4. Solve the resulting simultaneous equations to obtain the coordinates of the point.

From (2), $x = 3 - 3y$. Substituting into (1):
$2(3 - 3y) - y = 5$
$6 - 6y - y = 5 \implies 7y = 1 \implies y = \frac{1}{7}$
$x = 3 - 3\left(\frac{1}{7}\right) = \frac{18}{7}$
A point on the line is $\left(\frac{18}{7}, \frac{1}{7}, 0\right)$.

5. State the vector equation of the line $l$.

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

1. State the direction vector of $l$ and the condition for it to be parallel to $\pi_3$.

From part (b), the direction vector of $l$ is given by the cross product of the normal vectors of $\pi_1$ and $\pi_2$:
$\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 3 \\ -a \end{pmatrix} = \begin{pmatrix} a - 6 \\ 2a + 2 \\ 7 \end{pmatrix}$

The normal vector of $\pi_3$ is $\mathbf{n}_3 = \begin{pmatrix} b \\ -2 \\ 4 \end{pmatrix}$.
If $l$ is parallel to $\pi_3$, then $l$ is perpendicular to the normal of $\pi_3$, which means their scalar product must be zero ($\mathbf{d} \cdot \mathbf{n}_3 = 0$).

$\begin{pmatrix} a - 6 \\ 2a + 2 \\ 7 \end{pmatrix} \cdot \begin{pmatrix} b \\ -2 \\ 4 \end{pmatrix} = 0$

2. Evaluate the scalar product and factorise to obtain the given expression.

$b(a - 6) - 2(2a + 2) + 4(7) = 0$
$ab - 6b - 4a - 4 + 28 = 0$
$ab - 6b - 4a + 24 = 0$

Factorising by grouping:
$b(a - 6) - 4(a - 6) = 0$
$(a - 6)(b - 4) = 0 \quad \text{(shown)}$

1. Establish the base condition for a triangular prism using the result from (c).

For the three planes to form a triangular prism, they cannot intersect at a unique point. This requires the line of intersection of $\pi_1$ and $\pi_2$, $l$, to be parallel to $\pi_3$.
From part (c), this condition implies $(a - 6)(b - 4) = 0$, so $a = 6$ or $b = 4$.

2. Apply the non-parallel condition to $\pi_1$ and $\pi_3$ to deduce $a$.

The question states that all planes are non-parallel.
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 $b = 4$, then $\mathbf{n}_3 = \begin{pmatrix} 4 \\ -2 \\ 4 \end{pmatrix} = 2\mathbf{n}_1$, which means $\pi_1$ and $\pi_3$ are parallel.
Since the planes must be non-parallel, $b \neq 4$.
Therefore, we must have $a = 6$.

3. Apply the non-parallel condition to $\pi_2$ and $\pi_3$ to restrict $b$.

Substituting $a = 6$, the normal vector for $\pi_2$ is $\mathbf{n}_2 = \begin{pmatrix} 1 \\ 3 \\ -6 \end{pmatrix}$.
For $\pi_2$ and $\pi_3$ to be parallel, there must exist a scalar $k$ such that $\mathbf{n}_3 = k\mathbf{n}_2$.
Comparing the $y$ and $z$ components: $-2 = 3k \implies k = -\frac{2}{3}$, and $4 = -6k \implies k = -\frac{2}{3}$.
Then $b = k(1) = -\frac{2}{3}$.
To ensure $\pi_2$ and $\pi_3$ are non-parallel, $b \neq -\frac{2}{3}$.

4. Ensure the three planes have no point in common by substituting a point on $l$ into $\pi_3$.

For the planes to have no point in common, the line $l$ must not lie on $\pi_3$.
We find a point on $l$ by solving the equations for $\pi_1$ and $\pi_2$ with $a = 6$. Setting $z = 0$:
$2x - y = 5$
$x + 3y = 3 \implies x = 3 - 3y$
Substituting into the first equation:
$2(3 - 3y) - y = 5 \implies 6 - 7y = 5 \implies y = \frac{1}{7}$
$x = 3 - 3\left(\frac{1}{7}\right) = \frac{18}{7}$
So the point $\left( \frac{18}{7}, \frac{1}{7}, 0 \right)$ lies on $l$.
Substituting this point into the equation of $\pi_3$:
$b\left(\frac{18}{7}\right) - 2\left(\frac{1}{7}\right) + 4(0) \neq 3$
$\frac{18b - 2}{7} \neq 3 \implies 18b - 2 \neq 21 \implies 18b \neq 23 \implies b \neq \frac{23}{18}$