Finding the equation of plane that is transformed to a line

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 $x-y=0$?

-
I'll take a guess. The right hand side of your third displayed equation should be $[{t\atop t}]$. The solution set should be of the form $u{\bf v_1}+t{\bf v_2}$ ($u$ will be the free variable of the $2x3$ matrix). – David Mitra Nov 14 '11 at 17:38
The third display should not have the $t$. It was a typo. I have edited it. Thanks! – xenon Nov 14 '11 at 17:40
Maybe you should see what happens to the coordinates. $T\pmatrix{x\\y\\z} = \pmatrix{x+y-z\\z-x} =\pmatrix{t\\t}$. Start by replacing $z-x = m$, then you get $\pmatrix{y-m\\m}=\pmatrix{t\\t}$. – user13838 Nov 14 '11 at 17:43
@percusse ohh...thank you!! – xenon Nov 14 '11 at 18:01
@xEnOn No problem at all! – user13838 Nov 14 '11 at 18:05

Denote the points in ${\mathbb R}^3$ by $(z_1,z_2,z_3)$ and the points in the plane ${\mathbb R}^2$ by $(x,y)$. When $T:\ (z_1,z_2,z_3)\mapsto(x,y)$ then by the rules of linear algebra
$$\left[\matrix{x\cr y\cr}\right]= \left[\matrix{1 &-1&1 \cr -1&0&1\cr}\right]\ \left[\matrix{z_1\cr z_2\cr z_3}\right] = \left[\matrix{z_1-z_2+z_3\cr -z_1 +z_3\cr}\right]\ .$$
Now the question is: Which equation must the $z_k$ satisfy such that $x-y=0\$? As $x-y=2z_1-z_2$ it follows that $x-y=0$ iff $2z_1-z_2=0$; and the last equation is the equation of the plane in ${\mathbb R}^3$ that gets mapped onto the line $x=y$.