How to find the equation of plane given 3 points: $(a,0,0), (0,b,0),(0,0,c)$? 
Find the equation of plane given 3 points: $(a,0,0), (0,b,0), (0,0,c)$.

The way I would go about this is first finding two vectors:
$$
\overrightarrow{AB}=(0,b,0)-(a,0,0)=\langle -a,b,0\rangle\\
\overrightarrow{AC}=(0,0,c)-(a,0,0)=\langle -a,0,c\rangle
$$
Then we can get the normal vector by doing a cross product:
$$
\overrightarrow{AB}\times\overrightarrow{AC}=\langle -bc,-ac,-ab\rangle
$$
so the plane equation is something like:
$$
-bcx-acy-abz+d=0
$$
plug in one of the points for example $(a,0,0)$ to find $d$:
$$
-bcx-acy-abz+abc=0
$$

What I came across is another plane equation for these points:
$$
\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1
$$
which is inarguably much simpler.
How does one arrive to this representation?

Divide your equation by $abc$.
