I'd like to know, how can i prove it?
The vector equation of a plane that contains a point $P_0 = (x_0,y_0,z_0)$ and has normal vector $ \vec{n}=\vec{u} \times \vec{v}$, is given by
$$(x,y,z)=(x_0,y_0,z_0) + s\vec{u}+t\vec{v}$$
where s,t $\in \mathbb{R}$.
I tried using the "ordinary" way:
$$((x,y,x)-(x_0,y_0,z_0))\vec{n}=0$$
And this returned an equation that doesn't link to the first equation.
What you have is called a parametric equation for the plane. Each value of the two parameters $s, t$ gives a different point on the plane, and every point on the plane is given by some value for $s$ and $t$.
Where as the "ordinary way" is an equation in $(x,y,z)$ for which the points that satisfy it are exactly those on the plane.
Let $\vec r = (x,y,z)$ and for convenience i'll drop the subscript from $P_0$, just calling it $P$. The plane we are after is the one through $P$ and normal to a non-zero vector $\vec n$. That is, any direction along the plane will be perpendicular to $\vec n$. Thus $\vec r$ is on the plane if and only if $\vec r - P$ points along the plane, which holds if and only if it is perpendicular to $\vec n$. I.e., $$\vec n\cdot(\vec r - P) = 0$$
This is a test on the points $\vec r$ that is satisfied if and only if $\vec r$ is a point on the plane.
Now we are also given two vectors $\vec u, \vec v$ such that $\vec n = \vec u \times \vec v$. Now from the properties of the cross-product, this tells us a couple things. First, $\vec u$ and $\vec v$ are not parallel or anti-parallel, because if they were, then $\vec n$ would be $0$, and so could not be the normal vector to a plane. Second, $\vec u$ and $\vec v$ are normal to $\vec n$. That is, $$\vec n \cdot \vec u = 0\\\vec n \cdot \vec v = 0$$
For an arbitrary $s, t$, consider the point $P + s\vec u + t\vec v$. Is it on the plane? To find out, we plug it into the place of $\vec r$ in our test equation:
$$\begin{align}\vec n\cdot((P + s\vec u + t\vec v) - P) &= \vec n\cdot(s\vec u + t\vec v) \\&= \vec n\cdot(s\vec u) + \vec n\cdot(t\vec v) \\&= s(\vec n\cdot \vec u) + t(\vec n\cdot \vec v) \\&= s(0) + t(0) \\&= 0\end{align}$$
So, every point of the form $P + s\vec u + t\vec v$ is a point on the plane. The other side is to show that every point on the plane can be expressed as $P + s\vec u + t\vec v$ for some scalar values $s, t$. How exactly this can best be shown depends on what you currently know about vectors. So I am having to guess here.
Suppose $\vec r$ is an arbitrary point on the plane. And let $\vec w = \vec r - P$. Then we know from the test equation that $\vec w \cdot \vec n = 0$. Consider the following set of 3 equations in 3 unkowns $s, t, q$: $$u_xs + v_xt + n_xq = w_x\\u_ys + v_yt + n_yq = w_y\\u_zs + v_zt + n_zq = w_z$$ The determinant of this system turns out to be $\vec n \cdot(\vec u \times \vec v) = \vec n \cdot \vec n = \|\vec n\|^2$, which is not $0$. So the system has a unique solution $s, t, q$. But the system can be expressed vectorially as $$s\vec u + t\vec v + q\vec n = \vec w$$ Further, $$\begin{align}0 &= \vec n \cdot \vec w \\&= s(\vec n \cdot \vec u) + t(\vec n \cdot \vec v) + q(\vec n \cdot \vec n) \\&= 0 + 0 + q(\vec n \cdot \vec n)\end{align}$$ and so $q = 0$. Thus $\vec w = s\vec u + t\vec v$. I.e., $\vec r = P + s\vec u + t\vec v$.
So a point $\vec r$ is on the plane if and only if there are $s, t$ such that $\vec r = P + s\vec u + t\vec v$.