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?
 A: Divide your equation by $abc$.
A: You can arrive at this nice answer without much algebra if you know that the equation
$$
x + y + z = 1
$$
describes the plane containing the points $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ - the ends of the three unit coordinate vectors. Then you just scale the axes by $1/a$, $1/b$ and $1/c$ to get the equation you want.
Of course that works only of none of $a$, $b$ or $c$ is $0$, while your answer is always right.
Scaling axes is often a good approach to this kind of problem, so a technique worth remembering. It's a good way to turn circles into ellipses in the plane, with analogues is higher dimensions.
A: Anther way is to note that  a plane that does not passes thorough the origin has an equation of the form:
$$mx+ny+pz=1
$$
and substituting the point $(a,0,0)$ this gives $m=\frac{1}{a}$ , and the same for the other points.
A: Using general form of plane :$$Ax+By+Cz=D\\$$now put those points into it :
$$(a,0,0) \to Aa+0+0=D \to A=\frac{D}{a}\\
(0,b,0) \to 0+Bb+0=D \to B=\frac{D}{b}\\
(0,0,c) \to 0+0+Cc=D \to C=\frac{D}{c}\\$$ now put $A,B,C$ in the plane equation
$$Ax+By+Cz=D\\
\frac{D}{a}x+\frac{D}{b}y+\frac{D}{c}z=D$$ simplify $D$ 
you will have 
$$\frac{1}{a}x+\frac{1}{b}y+\frac{1}{c}z=1\\
\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$$
