Assume the plane:
$$ax+by+cz+d=0$$
Whose vector is gonna be:
$$\mathbf{\hat n}=\frac{(bc, ac, ab)}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}$$
Like in this picture:
How to find the expression for the area element, i.e. $d\mathbf{a}$?
I am confused, because all differentials are constrained to eachother:
$$a\ dx + b\ dy + c\ dz = 0$$
Can anyone help me?
On
Your normal vector is wrong. If the plane is implicitly given by an equation of the form $ax+by+cz=d$ with $(a,b,c)\ne{\bf 0}$ then $${\bf n}=\pm{1\over\sqrt{a^2+b^2+c^2}}(a,b,c)\ .$$ The scalar area element ${\rm d}\omega$ remains as such until you decide to set up a particular parametric representation $$(u,v)\mapsto{\bf r}(u,v)$$ of your plane. Then you obtain the pullback $${\rm d}\omega=|{\bf r}_u\times{\bf r}_v|\>{\rm d}(u,v)\ .\tag{1}$$ For example: If $c\ne0$ you could solve for $z$ and obtain the parametric representation $$(x,y)\mapsto\left(x,y,{d-ax-by\over c}\right)\ .$$ Compute $${\bf r}_x=(1,0,-a/c),\quad {\bf r}_y=(0,1,-b/c),\quad {\bf r}_x\times{\bf r}_y=(a/c,\>b/c,\>1)\ ,$$ and from $(1)$ you then get $${\rm d}\omega={\sqrt{a^2+b^2+c^2}\over |c|}\>{\rm d}(x,y)\ .$$
If $\{\hat{u},\hat{v},\hat{n}\}$ is an orthonormal base of $\mathbb{R}^3$, the area element of the plane is just $du\times dv$.
Anyway, in order to find the area of the triangle $ABC$ given by the points on our plane with non-negative coordinates, it is faster to compute the volume of $OABC$, that is (assuming $a,b,c>0$ and $d<0$) $\frac{-d^3}{6abc}$, and divide such volume by the distance of the plane from the origin. That gives one third of the area of $ABC$.