The planes $\Pi_1$ and $\Pi_2$ have vector equations
$\mathbf{r} = \lambda_1(\mathbf{i} + \mathbf{j} - \mathbf{k}) + \mu_1(2\mathbf{i} - \mathbf{j} + \mathbf{k}) \quad \text{and} \quad \mathbf{r} = \lambda_2(\mathbf{i} + 2\mathbf{j} + \mathbf{k}) + \mu_2(3\mathbf{i} + \mathbf{j} - \mathbf{k})$
respectively. The line $l$ passes through the point with position vector $4\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}$ and is parallel to both $\Pi_1$ and $\Pi_2$.

1. Find the normal vectors of planes $\Pi_1$ and $\Pi_2$.

Let $\mathbf{n}_1$ and $\mathbf{n}_2$ be the normal vectors to $\Pi_1$ and $\Pi_2$ respectively.
$\mathbf{n}_1 = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \times \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1(1) - (-1)(-1) \\ -1(2) - 1(1) \\ 1(-1) - 1(2) \end{pmatrix} = \begin{pmatrix} 0 \\ -3 \\ -3 \end{pmatrix}$
We can take a parallel normal vector as $\mathbf{m}_1 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$.

$\mathbf{n}_2 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \times \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 2(-1) - 1(1) \\ 1(3) - 1(-1) \\ 1(1) - 2(3) \end{pmatrix} = \begin{pmatrix} -3 \\ 4 \\ -5 \end{pmatrix}$

2. Find the direction vector of line $l$.

Since line $l$ is parallel to both $\Pi_1$ and $\Pi_2$, its direction vector $\mathbf{d}$ must be perpendicular to both normals. Thus, $\mathbf{d}$ is parallel to the cross product $\mathbf{m}_1 \times \mathbf{n}_2$.
$\mathbf{m}_1 \times \mathbf{n}_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \times \begin{pmatrix} -3 \\ 4 \\ -5 \end{pmatrix} = \begin{pmatrix} 1(-5) - 1(4) \\ 1(-3) - 0(-5) \\ 0(4) - 1(-3) \end{pmatrix} = \begin{pmatrix} -9 \\ -3 \\ 3 \end{pmatrix}$
We can simplify the direction vector by dividing by $-3$ to obtain $\mathbf{d} = \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}$.

3. Form the vector equation of line $l$.

The line passes through the point with position vector $4\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}$ and has direction vector $3\mathbf{i} + \mathbf{j} - \mathbf{k}$.
The vector equation is:
$\mathbf{r} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} + t \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}, t \in \mathbb{R}$

1. Find the direction of the line of intersection (building on part [0]).

Let $\mathbf{n}_1$ and $\mathbf{n}_2$ be the normal vectors to $\Pi_1$ and $\Pi_2$ respectively.
$\mathbf{n}_1 = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \times \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ -3 \\ -3 \end{pmatrix} = -3\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$
$\mathbf{n}_2 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \times \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} -3 \\ 4 \\ -5 \end{pmatrix}$
The direction vector $\mathbf{d}$ of the line of intersection $l'$ (and of $l$) is parallel to both planes, so:
$\mathbf{d} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \times \begin{pmatrix} -3 \\ 4 \\ -5 \end{pmatrix} = \begin{pmatrix} -9 \\ -3 \\ 3 \end{pmatrix} = -3\begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}$

2. Identify a point on the line of intersection $l'$ to formulate the shortest distance.

Observe from their parametric equations that $\Pi_1$ and $\Pi_2$ have no constant position vector term.
This means the origin $O(0, 0, 0)$ lies on both planes.
Thus, the line of intersection $l'$ passes through $O$, and has equation $\mathbf{r} = \mu \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}$.
The shortest distance between the parallel lines $l$ and $l'$ is equivalent to the perpendicular distance from the given point $A(4, 5, 6)$ on $l$ to the line $l'$.

3. Calculate the length of the projection of $\vec{OA}$ onto $l'$.

Let $\mathbf{a} = \vec{OA} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}$.
The length of the projection of $\vec{OA}$ onto the direction of $l'$ is:
$L = \frac{|\mathbf{a} \cdot \mathbf{d}|}{|\mathbf{d}|} = \frac{\left| \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix} \right|}{\sqrt{3^2 + 1^2 + (-1)^2}}$
$L = \frac{|12 + 5 - 6|}{\sqrt{11}} = \frac{11}{\sqrt{11}} = \sqrt{11}$

4. Use Pythagoras' theorem to find the perpendicular shortest distance.

Using Pythagoras' theorem on the right-angled triangle formed by $\vec{OA}$, its projection $L$, and the perpendicular distance $d$:
$d = \sqrt{|\mathbf{a}|^2 - L^2}$
$d = \sqrt{(4^2 + 5^2 + 6^2) - (\sqrt{11})^2}$
$d = \sqrt{77 - 11} = \sqrt{66}$