The transformation $A:\mathbb{R^4}\to \mathbb{R^4}$, represented by the matrix:
$A$ =$ \begin{bmatrix} 3 & 0 & 1 &0 \\ 0 & -3 & 0 & 1 \\ -1 & 0 & 1 & 0 \\ 0&-1&0&-1 \end{bmatrix} $ induces a projectivity $\varphi_A:\mathbb{P^3}\to \mathbb{P^3}$
consider the line $r = \begin{cases} x_0-x_1+x_3=0 \\ 2x_0-x_1-2x_2=0 \end{cases}$, $r\in \mathbb{P^3}$
How to find the equation of the line $s=\varphi_A^{-1}(r)$?
Could I say (?): let $P = [x_0, x_1, x_2, x_3]$ be the generic point on $\mathbb{P^3}$, $P \in \varphi_A^{-1}(r) \Leftarrow\Rightarrow \varphi_A(P)\in r.$ So I find $A(P)$ = $ \begin{bmatrix} 3x_0+x_2 \\ -3x_1+x_3 \\ -x_0+x_2 \\ -x_1-x_3 \end{bmatrix} $ and, replacing in $r$, I obtain: $s = \begin{cases} 3x_0+2x_1+x_2-2x_3=0 \\ 8x_0+3x_1-x_3=0 \end{cases}$
Thank you.
Your method works. Another method is to use $A$ directly: If you have a point transform given by the invertible homogeneous matrix $A$, i.e., $\mathbf p' = A\mathbf p$, then planes transform as $\mathbf\pi'=A^{-T}\mathbf\pi$ because $$\mathbf\pi^T\mathbf p = 0 \iff \mathbf\pi^T(A^{-1}\mathbf p') = (A^{-T}\mathbf\pi)^T\mathbf p'=0.$$
Your line is described as the meet of the planes $\mathbf\pi_1=[1:-1:0:1]^T$ and $\mathbf\pi_2=[2:-1:-2:0]^T$, so the inverse image of this line is the meet of $A^T\mathbf\pi_1 = [3:2:1:-2]^T$ and $A^T\mathbf\pi_2 = [8:3:0:-1]^T$.