The vector $n=(a,b,c)$ is the normal of the plane.
As the description of the plane using its normal gives no indication as to how its coordinate axes should be aligned, the conversions you ask for are not uniquely defined. To obtain any conversion matching your requirements, you first choose an arbitrary vector $v$ which is not a multiple of $n$. For simplicity, I'd use something like this:
$ v = \left\{\begin{array}{ll} (1, 0, 0) & \text{if } \lvert a\rvert\leq \lvert b\rvert, \lvert a\rvert\leq \lvert c\rvert \\ (0, 1, 0) & \text{if } \lvert b\rvert< \lvert a\rvert, \lvert b\rvert\leq \lvert c\rvert \\ (0, 0, 1) & \text{if } \lvert c\rvert< \lvert a\rvert, \lvert c\rvert< \lvert b\rvert \\ \end{array}\right. $
Now the cross product $a = n\times v$ will be non-zero and perpendicular to $n$, so it will be a vector in your plane. $b = n\times a$ will be perpendicular to both $n$ and $a$, so it will be a vector in the plane which is orthogonal to $a$. Normalize $a$ and $b$ (i.e. divide them by their length) and you have two orthogonal unit length vectors in your plane.
Let's call these vectors $a'$ and $b'$. Then every point $p$ in the plane can be written as $p = x'\cdot a' + y'\cdot b'$
So $(x', y')$ are the 2D coordinates of that point, while the linear combination above gives its 3D coordinates. This is the conversion you asked for. Conversely, the 2D coefficients can be computed from a 3D point $p$ using the dot product:
\begin{align*} x' &= p\cdot a' \\ y' &= p\cdot b' \end{align*}