Say I know a certain plane equation $ax+by+cz = d$, and i make $2$ new coordinate axis $v$ and $w$ (with $w$ is the output variable) that lies on that plane.
My question are :
1) Is it possible to make an equation of 2D curve without parametric variable (with the curve expressed in $v$ and $w$) ? .
2) How do you convert $v$ and $w$ back into $x,y,z$ ? (coordinate and equation). Do i need to know the angle made by the plane against $x,y,z$ plane ?
Sure! You can establish coordinates systems for any vector subspaces (case $d=0$) of $\mathbb{R}^3$. All you need is a basis for it!
Let's say your plane can be generates by $\{\vec{a} , \vec{b}\}$. Any point on it can be (uniquely) written as $u\vec{a} + v\vec{b}$ where $u,v$ are real numbers.
You can now work on your plane as you would in any two-dimensional space. For example, if $\vec{a}$ and $\vec{b}$ are unit-length and perpendicular (you can choose them pretty much however you want, so why not this way), you can write equations for curves in terms of $u$ and $v$ in a way that will sound familiar. For instance, a circle of center $(u_0, v_0)$ and radious $r$ on your plane would be written as $(u-u_0)^2+(v-v_0)^2=r^2$
Converting $(u,v)$ into $(x,y,z)$ is easy because $(x,y,z)=u\vec{a}+v\vec{b}$
If $d \neq 0$ things get a bit messier because $(0,0,0)$ would not lie on your plane, so you would need a new "origin" $(x_0, y_0, z_0)$ I would suggest you to apply the same ideas but then use the conversion $(x,y,z)=u\vec{a}+v\vec{b} + (x_0, y_0, z_0)$
To find our two vectors $\vec{a}, \vec{b}$, you just need to find three (as amd pointed correctly out, non-colinear) points on your plane, for instance, if $a,b,c,d \neq 0$, those could be $(d/a, 0, 0)$, $(0, d/b, 0)$, and $(0, 0, d/c)$. The differences between any two of those points are vectors parallel to your plane (so, you can use them for your basis)