Why does the moment of force (aka. torque) depend on the perpendicular distance? 
Couyld anyone explain how the lecturer concluded that $$(\underline{r_2} - \underline{r_1}) \times \underline{H} = \underline{p} \times \underline{H}$$
 A: Both $\vec p$ and $\vec r_2-\vec r_1$ start from the same point, and the tips of both vectors form line $H$. We know that the moment arm is the perpendicular distance from the axis of rotation to the force, and the perpendicular distance from the origin of both vectors to the line $H$ is obviously the same. 
You can also see that the cross product of two vectors
$\vec a \times \vec b$ = $|a||b|sin\theta$
has a $sin \theta$ term. In our case, this term finds the projection of $\vec r_2-\vec r_1$ on $\vec p$. (The projection is equal to the length of $\vec p$ so the moment arms are once again the same.)
Edit: By origin I mean the point where both $\vec p$ and $\vec r_2 - \vec r_1$ begin.
A: Because:
$$\vec p = \vec r_2 - \vec r_1 -r_3,$$
where $\vec r_3$ is the unlabelled side of the triangle that is parallel to $\vec H$.
$$ \vec r_2 - \vec r_1 = \vec p + \vec r_3$$
$$ (\vec r_2 - \vec r_1) \times \vec H = \vec p \times \vec H + \vec r_3 \times \vec H$$
but $\vec H$ and $\vec r_3$ are parallel so $\vec r_3 \times \vec H = 0$, and
$$ (\vec r_2 - \vec r_1) \times \vec H = \vec p \times \vec H$$.
A: A force acts on a line and picking any point along the line makes no difference in terms of the moment produced. In fact, the cross product is equivalent to the area of the triangle formed by the force vector and the point O by which the moment of calculated. In the area of the triangle it is the height of the triangle (perpendicular distance) that is important and here is the same thing. 
Here since any two point along the two lines would suffice, the author chooses the special case of the minimal distance (perpendicular distance).
