Finding intersection of 2 planes without cartesian equations? 
The planes $\pi_1$ and $\pi_2$ have vector equations:
  $$\pi_1: r=\lambda_1(i+j-k)+\mu_1(2i-j+k)$$
  $$\pi_2: r=\lambda_2(i+2j+k)+\mu_2(3i+j-k)$$
  $i.$ The line $l$ passes through the point with position vector $4i+5j+6k$ and is parallel to both $\pi_1$ and $\pi_2$. Find a vector equation for $l$.  

This is what I know:
$$l \parallel \pi_1, \pi_2 \implies l \perp n_1, n_2 \implies d = n_1\times n_2;\quad l:r=(4i+5j+6k)+\lambda d$$
However, that method involves 3 cross-products, which according to the examination report was an 'expeditious' solution which was also prone to 'lack of accuracy'. Rightfully so, I also often make small errors with sign changes leading me to the wrong answer, so if there is a more efficient way of working I'd like to know what that is.
Q1 How can I find the equation for line $l$ without converting the plane equations to cartesian form?

$ii.$ Find also the shortest distance between $l$ and the line of intersection of $\pi_1$ and $\pi_2$.

Three methods are described to solve the second part. However, I didn't understand this one:  

The determination of the shortest distance was perceived by most to be the shortest distance of the given point $4i+5j+6k$ to the line of intersection of the planes. Thus they continued by evaluating the vector product $(4i+5j+6k)\times (3i+j-k)$ to obtain $-11i+22j-11k$. The required minimum distance is then given immediately by $p = \frac{|-11i+22j-11k|}{\sqrt{11}} = \sqrt{66}$  

Q2 Why does the cross-product of the given point and the direction vector of the line of intersection of the planes get to the correct answer?
 A: 1) One method you can use to find line $L_1$ is to make equations in $x, y, z$ for each plane and solve them, instead of finding the cross product:
$$x=\lambda_1+2\mu_1, y=\lambda_1-\mu_1, z=-\lambda_1+\mu_1$$
$$x=\lambda_2+3\mu_2, y=\lambda_2+\mu_2, z=\lambda_2-\mu_2$$
What we'll do is find the line of intersection; this line is clearly parallel to both planes so we can use its direction for the line through $(4,5,6)$ that we want.
Since the line of intersection has to lie on both planes, it will satisfy both sets of equations. So equating them,
$$\lambda_1+2\mu_1 = \lambda_2+3\mu_2$$
$$\lambda_1-\mu_1 = 2\lambda_2+\mu_2$$
$$-\lambda_1+\mu_1 = \lambda_2-\mu_2$$
Adding the last two equations, we get,
$$\lambda_2 = 0$$
So,
$$x=3\mu_2, y=\mu_2, z=-\mu_2$$
This is a parametrisation of the solution which is the line. So obviously $L_1$ must be:
$$L_1: (4,5,6) + \mu_2(3,1,-1)$$
2) Notice that $(0,0,0)$ lies on the line of intersection $L_2$. So the difference vector $v$ between the point $(4,5,6)$ on the first line and $(0,0,0)$ on the second is $v=(4,5,6)$. The distance is really the length of the projection of $v$ on the perpendicular joining the two lines. This is $|v|\sin \theta$ where $\theta$ is the angle between line and plane.
Expressing $|v|\sin\theta$ as $$|v|\sin\theta=\frac{|v||w|\sin\theta}{|w|} = \frac{|v\times w|}{|w|}$$ (where $w$ is the direction vector of $L_2$) gives you the answer.
Here taking the cross product of the position vector of the given point with the direction vector of $L_2$ depended on $(0,0,0)$ lying on $L_2$; otherwise you would need to take the difference vector of the given point and any point on $L_2$.
