Is this a valid proof of this math challenge problem? From a fixed point P not in a given plane, three mutually perpendicular line segments are drawn terminating in the plane. Let a, b, c denote the lengths of the three segments. Show that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$$ has a constant value for all allowable configurations.
Here's my proof: 
Vector (a,b,c) can be used to represent the plane. Note that vectors with different magnitudes but equivalent directions represent parallel planes such that plane (xk,yk,zk) is parallel to (x,y,z). Let there be a vector (h,i,j) s.t it is equal to vector (x,y,z). No operations can be performed on (x,y,z) s.t x=/=h, y=/=i, z=/=j. Q.E.D.
This is my first attempt at doing very clever proofs, so please let me know if this is not a sufficient proof.
My proof is a one by contradiction. Basically it's saying that there is nothing you can do to vector (x,y,z) to make it different and still equivalent to another vector. The sentence before that on magnitudes is to emphasize that even though vectors have the same direction, without the same magnitudes they simply represent planes that are parallel to each other.
 A: As noted in the comments, the problem has been misread.
Here's an (inelegant) solution the problem. This is one of those sad coordinate-geometry arguments that illuminates very little about the geometry going on here, but can be done with bullet-headed algebraic computation.
If the statement of the problem is true, the constant value must be:
$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{d^2}\tag{1}$$
where $d$ is the distance of the point $P$ to the plane. This is because you can rotate your frame of vectors $\mathbf a,\mathbf b,\mathbf c$ so that only $\mathbf a$ is on the plane and $\mathbf b,\mathbf c$ are essentially infinite. So the constant value, if it exists, has to be $\frac{1}{d^2}$, where $d$ is the distance from $P$ to the plane. 
(While the problem specifically states that $\mathbf b,\mathbf c$ must be on the plane, what this really amounts to is rotating the frame so that $\mathbf a$ is close to the nearest point to $p$, and $\mathbf b,\mathbf c$ are very far away.)
So another way to state this theorem is to say that if you have a plane in $\mathbb R^3$ which is distant $d$ from the origin and contains $(a,0,0),(0,b,0),$ and $(0,0,c)$, then $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{d^2}.$$
Let $\mathbf v=(x_0,y_0,z_0)$ be the point on the plane closes to $(0,0,0)$. Then the plane can be written as the set of $(x,y,z)$.
$$\mathbf v\cdot(x,y,z)=\mathbf v\cdot \mathbf v=d^2$$ 
In particular, then $$\mathbf v\cdot(a,0,0)=ax_0=d^2\\
\mathbf v\cdot(0,b,0)=by_0=d^2\\
\mathbf v\cdot(0,0,c)=cz_0=d^2$$
  So:
$$\frac{1}{a^2} = \frac{x_0^2}{d^4}\\
\frac{1}{b^2} = \frac{y_0^2}{d^4}\\
\frac{1}{c^2} = \frac{z_0^2}{d^4}$$
So:
$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{x_0^2+y_0^2+z_0^2}{d^4}=\frac{1}{d^2}.$$
Alternative approach: Given $a,b,c\neq 0$, the plane through $(a,0,0),(0,b,0),(0,0,c)$ in $\mathbb R^3$ has equation:
$$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$$
Find the nearest point to the origin using this equation, and show $(1)$.
A useful theorem:

The nearest point to the origin on the plane $$\alpha x+\beta y+\gamma z=C$$
  is a scalar multiple of $(\alpha,\beta,\gamma)$, specifically:
  $$(x,y,z)=\frac{C}{\alpha^2+\beta^2+\gamma^2}(\alpha,\beta,\gamma)$$

Then we see that $d^2=x^2+y^2+z^2=\frac{C^2}{\alpha^2+\beta^2+\gamma^2}$.
