Suppose I have a transformation $T:\mathbb{R}^3\rightarrow \mathbb{R}^2$ and its matrix: $$T=\begin{bmatrix} 1 & -1 & 1\\ -1 & 0 & 1 \end{bmatrix}$$
I am told that there is a plane being transformed to a line of the equation $x-y=0$. And I want to find out the equation of the plane.
I thought I could first express the plane in its explicit form: $$\begin{bmatrix} x\\ y \end{bmatrix}=t \begin{bmatrix} 1\\ 1 \end{bmatrix} ,t\in\mathbb{R}$$ And then make the transformation equate to it: $$\begin{bmatrix} 1 & -1 & 1\\ -1 & 0 & 1 \end{bmatrix}\begin{bmatrix} x_{plane}\\ y_{plane}\\ z_{plane} \end{bmatrix}= \begin{bmatrix} 1\\ 1 \end{bmatrix}$$ Solve for its solution space: $$\begin{bmatrix} x_{plane}\\ y_{plane}\\ z_{plane} \end{bmatrix}= t\begin{bmatrix} 1\\ 2\\ 1 \end{bmatrix}+ \begin{bmatrix} -1\\ -2\\ 0 \end{bmatrix},t\in\mathbb{R}$$
But this is obviously wrong because first, this is a line and second, this is not an implicit equation of the plane.
What should I do to get the equation(implicit form) of the plane that is transformed to the line $x-y=0$?