Parallel planes of lattice and reciprocal vector (I edited my message to ask one question at a time to simplify)
My question is the following.
A vector $\overrightarrow{G}$ is a vector of the reciprocal lattice if it satisfies : $ \forall \overrightarrow{r} \in L~ ~ e^{i \overrightarrow{G}.\overrightarrow{r}}=1 $ where $L$ means "lattice".
We also say that parallel planes of a lattice are orthogonal to a vector $\overrightarrow{G}$.
But I don't totally understand this.
Indeed, imagine that we have a plane of the lattice orthogonal to the $ \overrightarrow{U_z}$ axis (a crystal plane).
For any node of the Bravais lattice localised by $r_n$ of this plane, all the vectors of the reciprocal lattice verifies : $\overrightarrow{G}.\overrightarrow{r_n}=2 \pi m$ where $ m \in \mathbb{Z} $.
But it doesn't mean that it exists a vector $\overrightarrow{G}$ that verifies $\overrightarrow{G}.\overrightarrow{r_n}=2 \pi m$ with the same $m$ for all the points $\overrightarrow{r_n}$ of this plane. It exists such a vector only if it is orthogonal to this crystal plane.
So, first question:
When we have a set of parallel planes of a lattice, how can we know that it exists a reciprocal vector orthogonal to these planes ?
I know that I could for any lattice write the reciprocal vectors using the formulas that give basis of reciprocal lattice for 2D or 3D crystals for example but I am expecting a more "general" answer to better understand it.
Also if possible I would like to avoid the notion of notation of planes (hkl) because I haven't seen it yet and I don't think it is needed to understand my question. But if you are sure you can't explain it without it then go on :) 
 A: I'm not sure your definition of reciprocal vector is exact. A vector $\vec{G}$ belongs to the reciprocal lattice if it satisfies
$$ exp[i\vec{G}\cdot\vec{R}]=1 $$ for any $\vec{R}$ that belongs to the Bravais lattice of the crystal. So, taking your plane orthogonal to the $\vec{U_Z}$ axis, for any point in the plane localised by the $\vec{R}$ indeed you have a relation of the kind 
$$ \vec{G}\cdot\vec{R} = 2\pi m \, ,\, m\in \mathbb{Z}$$ but it doesn't mean that all the points of the plane satisfies that relation. The important thing to keep in mind is that the reciprocal vectors don't exist in the real space, but they live in their own space, the reciprocal space. In fact, the $\vec{G}$ have the dimension of an inverse of a lenght: $[\vec{G]}=[L]^{-1}$; thus, the picture of an actual vector is misleading.
EDIT
The answer to your question is that the periodicity of your Bravais lattice ensures you that its reciprocal lattice exist, because it is its Fourier transform. Actually, one lattice is the Fourier transform of the other.
A: For any crystalographic plane it is possible to take a parallel one passing by the origin.
In this plane, we can select 2 basis vectors $\mathbf b$ and $\mathbf c$ that generate all sites of the plane: $\mathbf r = n_2\mathbf b +  n_3\mathbf c$
Another vector $\mathbf a$, not in this plane, can be chosen to complete a basis for the Bravais lattice. It is important to remark that such basis are not unique.
It is possible to define a basis for the reciprocal lattice from the basis ($\mathbf a,\mathbf b,\mathbf c$) of the Bravais lattice :
$$\mathbf G_1 = \frac{(2\pi) \mathbf b \times \mathbf c}{(\mathbf a \times \mathbf b).\mathbf c}$$
$$\mathbf G_2 = \frac{(2\pi) \mathbf c \times \mathbf a}{(\mathbf a \times \mathbf b).\mathbf c}$$
$$\mathbf G_3 = \frac{(2\pi) \mathbf a \times \mathbf b}{(\mathbf a \times \mathbf b).\mathbf c}$$
$\mathbf G_1.\mathbf b =  \mathbf G_1.\mathbf c = 0$, because $\mathbf b \times \mathbf c$ is normal to each of that vectors.
$$\mathbf G_1.\mathbf a = \frac{2\pi*(\mathbf b \times \mathbf c).\mathbf a}{(\mathbf a \times \mathbf b).\mathbf c} =  2\pi$$ because the scalar triple product is invariant for cycling permutations. The same procedure for the remaining vectors.
$\mathbf G_1$ is normal to $\mathbf b$ and $\mathbf c$, and that way normal to the plane, as required.
