Find equation of the plane that this $x=\frac{1+t}{1-t}, y=\frac{1}{1-t^2}, z=\frac{1}{1+t}$ curve lies 
Prove that all points of the given curve lie in one plane, and find
the equation of that plane:
$$x=\frac{1+t}{1-t}, y=\frac{1}{1-t^2},  z=\frac{1}{1+t}.$$

If the given curve lies in one plane, then
$$a\left(\frac{1+t}{1-t}\right)+b\left(\frac{1}{1-t^2}\right)+c\left(\frac{1}{1+t}\right)+d=0.$$
Solving this I get $2a=c,a=d,a=\frac{-b}{4}.$
How from this find equation of plane? Or maybe I did something wrong?
When putting values back into equation of plane I get
$$ax-4ay+2az+a=0.$$ Now, problem is I can't cancel $a$ here because first of all, I need to prove that such plane exists.

Observe that (for $t\neq\pm1$)
$$\begin{align}
x&=\frac{2}{1-t}-1\iff\frac{1/2}{1-t}=\frac{x+1}{4},\\
y&=\frac{1/2}{1-t}-\frac{1/2}{1+t},\\
z&=\frac{1}{1+t}\iff\frac{1/2}{1+t}=\frac{z}{2}.
\end{align}$$
From here
$$y=\frac14(x+1)-\frac12z\iff x-4y+2z=-1.$$
Alternatively, the curves equation is
$$\begin{pmatrix}
-1\\ 0\\ 0
\end{pmatrix}+
\frac{1}{1-t}
\begin{pmatrix}2\\ 1/2 \\0
\end{pmatrix}+
\frac{1}{1+t}
\begin{pmatrix}
0\\-1/2\\1
\end{pmatrix},
$$
clearly an equation of a part of a plane.
