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.
 A: We have the following parameteric curve
$x(t) = \dfrac{1 + t}{1 - t} , y(t) = \dfrac{1}{1 - t^2} , z(t) = \dfrac{1}{1 + t} $
As the OP pointed out, we can simply assume that the curve lies on the plane
$$ a x + b y + c z + d = 0 $$
And then try and find $a,b,c,d$.  Using this method, and multiplying through by  $(1 - t^2) $ results in
$ a (1 + t)^2 + b + c (1 - t) + d (1 - t^2) = 0 $
Collecting terms containing $t^2 , t $ and the constant terms we get
$ t^2 ( a - d ) + t ( 2 a - c ) + ( a + b + c + d) = 0 $
Taking $d = 1 $ , then
$ a = 1, c = 2 , b = -4 $
Thus this curve does indeed lie on the plane
$ x - 4 y + 2 z + 1 = 0 $
which has been pointed out by @RKK
A more involved method that is not necessarily better, is to differentiate $(x(t), y(t), z(t))$ to obtain
$T(t) = (x'(t), y'(t), z'(t) ) = \left( \dfrac{ 2 }{(1 - t)^2 }, \dfrac{2 t}{(1 - t^2)^2} , -\dfrac{1}{(1 + t)^2 } \right) $
And then differentiate $T(t)$ to obtain
$T'(t) = ( \dfrac{ 4 }{ (1 - t)^3 } , \dfrac{ 2  + 6 t^2 }{ (1 - t^2)^3 } , \dfrac{ 2 }{(1 + t)^3 }) $
Now, we find the cross product $T(t) \times T'(t) $
$ T(t) \times T'(t) = \begin{vmatrix} \mathbf{i} && \mathbf{j} && \mathbf{k} \\
 \dfrac{ 2 }{(1 - t)^2 } && \dfrac{2 t}{(1 - t^2)^2} && -\dfrac{1}{(1 + t)^2 } \\
 \dfrac{ 4 }{ (1 - t)^3 } && \dfrac{ 2  + 6 t^2 }{ (1 - t^2)^3 } && \dfrac{ 2 }{(1 + t)^3 } \end{vmatrix} $
$ = \mathbf{i} \left( \dfrac{ 4 t }{ (1 - t^2)^2 (1 + t)^3 } + \dfrac{ 2 + 6 t^2 }{ (1 + t)^2 (1 - t^2)^3 } \right) \\
 + \mathbf{j} \left( - \dfrac{4}{ (1 + t)^2 (1 - t)^3 } - \dfrac{4}{(1 - t)^2 (1 + t)^3 } \right)
+ \mathbf{k} \left(  \dfrac{ 2 (2 + 6 t^2)}{(1 - t)^2 (1 - t^2)^3 } - \dfrac{ 8t }{ (1 - t)^3 (1 - t^2)^2} \right) $
And this simplifies to,
$ =\dfrac{2}{(1 - t^2)^3}  \left( \mathbf{i}  - 4 \mathbf{j} + 2 \mathbf{k} \right) $
Hence the vector $(1, -4, 2) $ is always normal to the curve, which means the curve lies in the plane $ x - 4 y + 2 z + d = 0 $
To find $d$ , substitute $t = 0$, you get the point $(1, 1, 1) $ on the plane, hence
$ d = - (1 - 4 + 2) = 1 $
Finally the equation of the plane is  $ x - 4 y + 2 z + 1 = 0 $
But the method of the OP and @RKK is much simpler than the second method that involved the tangent vector and its derivative.
A: I don't quite understand the added value of using  the cross product in the alternative. To prove that the curve lies in a plane with normal vector $\hat{n} = (a,b,c)$  it is sufficient that the tangent vector $\hat{\tau} = (x,_t, y,_t,z,_t) $ of the curve is perpendicular to $\hat{n}$ for $ \forall  \ t \ \epsilon \ \mathbb{R}_{\setminus \{-1,1\}}$ .
So,
\begin{align}
\hat{\tau} &= (x,_t, y,_t,z,_t) \\
&= \left(\frac{2}{(1-t)^2}, \frac{2t}{(1-t^2)^2}, -\frac{1}{(1+t)^2}\right)\\
\left\langle \hat{n} ,\hat{\tau}\right\rangle &= 0\\
&= a\frac{2}{(1-t)^2}+b\frac{2t}{(1-t^2)^2}-c\frac{1}{(1+t)^2}\\
&= 2a(1+t)^2+ 2bt -c(1-t)^2 \\
 &= 0\\
\text{for }& \quad t=0 \ : \quad 2a-c = 0\\
\Rightarrow a = 1 \quad &\wedge\quad  c= 2 \quad \text{(we don't care about the size of the normal vector)}
\end{align}
On the condition that $t\neq 0$ we get
\begin{align}
b &= -\frac{(1+t)^2 - (1-t)^2}{t}\\
&= -4
\end{align}
With this, the existence of a plane is proved. And to determine the displacement of the plane we just have to solve
$$x-4y+2z + d =0$$
for any allowed value of $t$.
A: 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.
A: 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.
