<s>[INST] 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.
 [/INST] 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.
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