Area of the polygon formed by cutting a cube with a plane I want to determine the area of a polygon that is formed when a cube is cut by a plane like shown below:

Where the blue triangle is in fact a part of the plane described by 
$$m_x x + m_y y +m_z z = \beta \tag{1} $$
and the box's dimensions are $1\times1\times1$. The polygon I am interested in is highlighted in red.
From $(1)$ I can easily determine the location of the points $r_i$ to be
$$r_1=\left(0,\frac{\beta-m_z}{m_y},1\right); \;\; r_2=\left(\frac{\beta-m_z}{m_x},0,1\right); $$
$$r_3=\left(1,0,\frac{\beta-m_x}{m_z}\right); \;\; r_4=\left(1,\frac{\beta-m_x}{m_y},0\right); $$
$$r_5=\left(\frac{\beta-m_y}{m_x},1,0\right); \;\; r_6=\left(0,1,\frac{\beta-m_y}{m_z}\right); $$
I know I can determine the area of an irregular polygon with $N$ vertices located at $(a,b)$ (in some plane) as:
$$A=\frac{1}{2}\left(\sum_{i=1}^{N-1} a_i b_{i+1}-\sum_{j=1}^{N-1} a_{j+1} b_{i}+a_Nb_1-b_1a_N\right) \tag{2}$$
I somehow have to transform the coordinates $r_i$ such that they are in a plane where the third coordinate is 0, i.e. I have to align the polygon with one of the axis of my coordinate system and that is where I get stuck. How can I transform $r_i$ such that the plane aligns with my coordinate system and I can use Eq.$(2)$
P.S. I will also accept an answer with a completely different way of getting the area of the polygon, the method I describe is just the one I know of.
 A: Project your polygon $P$ to the $(x,y)$-plane by forgetting the $z$-coordinates of the vertices, and compute the area of  $P'$ using your formula for areas of plane polygons. Then
$${\rm area}(P)={1\over\cos\phi}{\rm area}(P')\ ,$$
where $\phi$ is the angle between the $z$-axis and and the normal to the cutting plane $(1)$.
A: This argument is specifically for the case in which the plane's intersection with the cube is a hexagon.

Let the plane meet the coordinate axes as $A = (a,0,0)$, $B = (0,b,0)$, $C = (0,0,c)$, so that its equation can be written
$$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$
(In terms of your notation, this means $a = \beta/m_x$, $b = \beta/m_y$, $c = \beta/m_z$.) Capitalizing your $r_i$ for notational consistency, the area you seek is
$$|\triangle ABC| - |\triangle AR_3R_4| - |\triangle BR_5R_6| - |\triangle CR_1R_2| \qquad(\star)$$
Note that each of the triangles referenced in $(\star)$ is the "hypotenuse-face" of a right-corner tetrahedron. Moreover, the tetrahedra are similar, so that the areas in question are related by (the squares of) appropriate proportions of the corresponding dimensions measured along coordinate axes. For instance, with $O = (0,0,0)$ and $I = (1,0,0)$, we have
$$|\triangle AR_3R_4| = \left(\frac{|AI|}{|AO|} \right)^2 |\triangle ABC| = \frac{\;(a-1)^2}{a^2}|\triangle ABC|$$
and likewise for $|\triangle BR_5R_6|$ and $|\triangle CR_1R_2|$, so that $(\star)$ becomes
$$|\triangle ABC|\left( 1 - \frac{(a-1)^2}{a^2} - \frac{(b-1)^2}{b^2} - \frac{(c-1)^2}{c^2}\right) \qquad (\star\star)$$
Now, to get $|\triangle ABC|$ itself, we simply apply the Pythagorean-like Theorem of de Gua for right-corner tetrahedra, which gives the square of the hypotenuse-area as the sum of the squares of the (easily-computed) leg-areas:
$$|\triangle ABC|^2 = |\triangle OBC|^2 + |\triangle OCA|^2 + |\triangle OAB|^2 = \frac{1}{4}\left( b^2c^2 + c^2 a^2 + a^2 b^2 \right)$$
The makes $(\star\star)$ equal to
$$\frac{1}{2}\sqrt{ b^2c^2 + c^2 a^2 + a^2 b^2}\;\left( 1 - \frac{(a-1)^2}{a^2} - \frac{(b-1)^2}{b^2} - \frac{(c-1)^2}{c^2}\right)$$
$$= \frac{\beta^2 \sqrt{m_x^2 + m_y^2 + m_z^2}}{2m_x m_y m_z} \frac{1}{\beta^2}\left(-2 \beta^2 + 2 \beta (m_x+m_y+m_z) - ( m_x^2 + m_y^2 + m_z^2)\right)$$
$$= \frac{\sqrt{m_x^2 + m_y^2 + m_z^2}}{2m_x m_y m_z}\left(-2 \beta^2 + 2 \beta (m_x+m_y+m_z) - ( m_x^2 + m_y^2 + m_z^2)\right) \qquad (\star\star\star)$$
This agrees with @achillehui's answer, under the assumption that $m_x^2 + m_y^2 + m_z^2 = 1$.
