Find the intersection of two planes. Find the intersection of the planes $x+(y-1)+z=0$ and $-x+(y+1)-z=0$. These two planes are 3-dimensional and I am confused on how to solve it.

Hint : You have to solve the system :
$$\left\{\begin{array}{l}x+(y-1)+z=0\\-x+(y+1)-z=0  \end{array} \right. $$
that means expressing two of the variables in terms of the third one. Imagine you get something like :
$$\left\{\begin{array}{l}x=1+3z\\y=-2+z  \end{array} \right. $$
then adding $z=t$ you get a system of parametric equation of a line :
$$\left\{\begin{array}{l}x=1+3t\\ y=-2+t\\z=t  \end{array} \right., t\in\mathbb{R}.$$
in my example this would be the line passing through $(1;-2;0)$ (constant values in the equations) with direction $\overrightarrow{u}(3;1;1)$ (coefficients multiplying $t$ in the equations)
Your example is a bit particular as it yields $y=0$ quickly. You can write it as $y=0z$ and continue as planed.
