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)?
Let $\Pi_1, \Pi_2$ be the planes in the question.
Let $P_1$ the the plane $ax+by+cz=0$ (parallel to $\Pi_1$) and let $P_2$ be the plane $ex+fy+gz=0$ (parallel to $\Pi_2$).
Pick a point $d \in P_1$ that is not on $P_2$. Such a point must exist, otherwise the planes are parallel.
We see that $a d_1 + b d_2 + c d_3 = 0$ and $e d_1 + f d_2 + g d_3 \neq 0$.
Now pick any point $(x,y,z)$ on the plane $\Pi_1$ described by $ax+by+cz+d = 0$ and consider the point $(x,y,z) + t (d_1,d_2,d_3)$ as $t$ varies. For any $t$ this point lies in $\Pi_1$, and if we choose $t^* = - {ex + f y + g z+h \over e d_1 + f d_2 + g d_3}$, we see that the point corresponding to $t^*$ lies on $\Pi_2$.
Without loss of generality, we can take one plane to be the XY-plane, since the problem is fixed under translation and rotation. Let the second plane have normal vector $u=(u_1, u_2, u_3)$ and some point $p=(p_1, p_2, p_3)$. Noting that $v=(0, -u_3,u_2)$ is perpendicular to $u$, $p+tv$ is in that plane for all scalar $t \in \mathbb{R}$. Choosing $t = -p_3/u_2$ gives $(p_1, p_2+p_3u_3/u_2, 0)$ is in both planes, proving that they intersect.
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 $v\times w$.
It seems that this is not meant to be a problem about the axioms of euclidean geometry, but about 3d analytic geometry. If the two planes are not parallel then the two vectors ${\bf u}:=(a,b,c)$ and ${\bf v}:=(e,f,g)$ have a non-vanishing cross product ${\bf u}\times{\bf v}$. It follows that, e.g., $af-be\ne0$. Therefore the system of linear equations $$\eqalign{ax+by&=-cz-d\cr ex+fy&=-gz-h\cr}$$ has a unique solution $(x,y)$ for any choice of $z$. In other words, by solving this system we obtain $x$ and $y$ as functions of $z$: $$x=Az+B,\quad y=Bz+D\ ,$$ whereby $A$, $\ldots$, $D$ depend on the data $a$, $\ldots$, $h$. We then can use these functions in order to set up a parametric representation of the line $\ell$ (not "the point") of intersection of these two planes as follows: $$\ell: \quad z\mapsto\bigl(Az+B,Cz+D,\ z\bigr)\qquad(-\infty<z<\infty)\ .$$ Here we have used the $z$-coordinate of the moving point on $\ell$ as parameter.
"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."
THis is not a proof as analytic geometry was defined with the Euclid's 5th postulate incorporated into the entire basis of converting lines and planes to coordinate system. so this is circular.
"I wonder if there is a quicker and more elegant proof (without linear algebra --- this is high school level geometry)?
No. Because it isn't true. Read up on non-euclidean geometry.
Euclid's 5th postulate-- that if the interior angles of two line and an mutually intersecting third line are supplimentary than the lines never intersect, but if not, the lines will eventually intersect on the side where the interior angles are less than supplementary--- is an axiom.
If we modify the axiom to render that there are infinite number of lines that will not intersect with interior angles of varying measurements, or if we modify it so that all pairs of lines will always intersect twice even if the interior angles are supplimentary--- then the new axiom will yield different but equally valid systems of geometry.
The system of two equations in three unknowns
$$\begin{cases}ax+by+cz+d=0\\ex+fy+gz+h=0\end{cases}$$ usually has a simple infinity of solutions, as one of the unknowns can be set arbitrarily.
There is an exception when the linear coefficients are proportional to each other, $$\lambda a=\mu e, \lambda b=\mu f, \lambda c=\mu g.$$
Then by combining the equations, one gets
$$\lambda d=\mu h.$$
There are two cases:
this equality holds: then the two original equations have the same solution set, which is a plane; the given planes are merged.
the equality does not hold: then the system is impossible and has no solution; the given planes are parallel and distinct.