Proof that two non-parallel planes must intersect? I managed to find, by enumeration, the intersection point of two planes $ax+by+cz+d=0$ and $ex+fy+gz+h=0$, in all possible cases (with the condition that the planes are not parallel). But this is a very ugly proof.
I wonder if there is a quicker and more elegant proof (without linear algebra --- this is high school level (Euclidean) geometry)?

You want to show that if $v,w$ are linearly independent vectors in $\mathbb R^3$, then the $2\times 3$ matrix $A$ formed by putting $v$ and $w$ in two rows defines a map $A:\mathbb R^3\to \mathbb R^2$ that is onto. It suffices you show that the kernel of $A$ has dimension $1$ when $v,w$ are linearly independent, from this follows that the image of $A$ has dimension $2$; that is, we can always solve the system 
$$ v\cdot x=\lambda,w\cdot x=\mu $$
for any two scalars $\mu,\lambda$ (this is what you want). The claim that $A$ has one dimensional kernel is the same as saying two non parallel planes passing through the origin intersect in a line. Can you prove this?
To do this, you want to show that the simultaneous equations 
$$ v\cdot x =0,w\cdot x=0$$
have a solution set equal to a line. The fact that $w$ and $v$ are linearly independent means that they are not a scalar multiple of each other. This means the cross product $u=v\times w$ is nonzero, and this gives a nonzero vector $u$ that solves the above, so the kernel has dimension at least $1$ (this is always true!), so we have to check these are all the solutions. 
Pick another solution, $x$. Because $(v,w,v\times w)$ is a basis of $\mathbb R^3$, we can write $x$ as a linear combination of $v,w,v\times w$, and if $x\cdot w=x\cdot v=0$, then in fact $x$ is a multiple of $v\times w$ (take the inner product against $v,w$ to see the corresponding coefficients are zero). Thus, as desired, $\ker A $ is generated by $v\times w$. 
