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.

Partial answer
$\vec r_0:= (x(0),y(0),z(0))=(1,1,1);$
Equation of a plane passing through $\vec r_0,$ where  $\vec n:=(a, b, c)$ is the normal of the plane:
$\vec n \cdot (\vec r - \vec r_0)=$
$(a,b,c)\cdot (x-1,y-1,z-1)=$
$a(x-1)+b(y-1)+c(z-1)=0;$
If the curve lies in the plane then
$\vec r(t) =(x(t), y(t), z(t)) $ satisfies this equation for $t\not =\pm1$.
$ax(t) +by(t)+cz(t)=$
$a+b+c;$
$a(1+t)^2+b+c(1-t)=$
$(a+b+c)(1-t^2);$
A quadratic equation in $t$.
It is satisfied for all $t$ if the coefficients of $t^2,t$ and the constant are zero, which gives $3$ equations for $a, b, c.$
Assuming the system of equations is consistent and has one solution the curve lies in the plane.
