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.


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...
