Relationships between points lines and planes Develop the Cartesian equation of a plane with $x$-intercept $a$, $y$-intercept $b$ and $z$-intercept $c$. 
Show that the distance $d$ from the origin to this plane is given by $$\frac{1}{d^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$$ 
In the picture below I have included what I have done so far. I was able to set all my variables and begin to define them.

 A: Normal to the plane, using the position vectors of the intercepts given, is the vector $$\left(\begin{matrix}a\\-b\\0\end{matrix}\right)\times\left(\begin{matrix}a\\0\\-c\end{matrix}\right)=\left(\begin{matrix}bc\\ac\\ab\end{matrix}\right)$$
This vector is parallel to $$\left(\begin{matrix}\frac 1a\\ \frac 1b\\ \frac 1c\end{matrix}\right)$$
Now the distance from the origin to the plane is the projection of $\overrightarrow{OA}$ onto this normal, and $\overrightarrow{OA}\cdot\underline{n}=1$.
Therefore $$d=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}$$
And hence the result...
A: Let $A = (a,0,0)$, $B = (0,b,0)$, $C = (0,0,c)$. The normal to the plane is in the direction $N = (B-A)\times(C-A) = (bc, ca, ab)$. The plane passes through the point $A$, so its equation is $(X-A)\cdot N =0$. Note that $A \cdot N = abc$, so this can also be written as:
$$
bcx + cay + abz = abc
$$
The line through the origin normal to the plane has equation $L(t) = tN$. Where this line intersects the plane we get a point $F$. This is the "foot" of the normal from the origin to the plane, so it's the point of the plane that's closest to the origin. At $F$, we have $(tN - A) \cdot A = 0$. This gives $t = (A \cdot N)/(N \cdot N) = abc/(N \cdot N)$, so the intersection point is 
$$
F = \frac{abc}{N \cdot N}N = \frac{abc}{\| N \|^2}N
$$
The distance $d$ from the origin to $F$ is given by
$$
d = \|F\| = \frac{abc}{\| N \|^2}\|N\| = \frac{abc}{\| N \|}
= \frac{abc}{\sqrt{b^2c^2 + c^2a^2 + a^2b^2}}
$$
This formula is correct even if one of $a$, $b$, $c$ is zero and the other two are non-zero, in which case it gives $d=0$, as you would expect.
If $a$, $b$, $c$ are all non-zero, this formula gives
$$
\frac{1}{d^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}
$$
A: Another attempt, since the OP finds my first answer confusing ...
Let $A = (a,0,0)$, $B = (0,b,0)$, $C = (0,0,c)$. The normal to the plane is in the direction $N = (B-A)\times(C-A) = (bc, ca, ab)$. A unit vector in this direction is
$$
U = \frac{N}{\|N\|} = \frac{(bc,ca,ab)}{\sqrt{b^2c^2 + c^2a^2 + a^2b^2}}
$$ 
The distance $d$ from the origin to the plane is the length of the projection of $\vec{OA}$ onto the vector $U$. Since $U$ is a unit vector, this projected length is just  $A \cdot U$. So, we have
$$
d = A \cdot U = \frac{abc}{\sqrt{b^2c^2 + c^2a^2 + a^2b^2}}
$$
This formula is correct even if one of $a$, $b$, $c$ is zero and the other two are non-zero, in which case it gives $d=0$, as you would expect.
If $a$, $b$, $c$ are all non-zero, this formula can be rearranged to give
$$
\frac{1}{d^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}
$$
