We have a line in $ \Bbb R^3 $ given as intersetion of two planes: $$ \left\{ \begin{aligned} A_1x+B_1y+C_1z + D_1 &=0 \\ A_2x+B_2y+C_2z + D_2 &=0 \\ \end{aligned} \right. $$ How to represent it in parametric form: $$ \left\{ \begin{aligned} x &= x_0 +at\\ y &= y_0 +bt \\ z &= z_0 +ct \\ \end{aligned} \right. $$ ?
example I'm doing is:
$
l:
\left\{
\begin{aligned}
x + y - z + 1 &=0 \\
x - y + z - 1 &=0 \\
\end{aligned}
\right.
$
On
Row reduce the system of equations. Generically, you will get an equation with two variables and one with three. There is going to be a repeated variable in both, that is your parameter. Write the remaining variables in terms of this common variable and you are done.
In your example:
$ l: \left\{ \begin{aligned} x + y - z + 1 &=0 \\ x - y + z - 1 &=0 \\ \end{aligned} \right. $
Add the first equation to the last to get:
$ l: \left\{ \begin{aligned} x + y - z + 1 &=0 \\ x &=0 \\ \end{aligned} \right. $
We can take $y$ or $z$ as the parameter. For example let's take $z$. Then
$ \begin{aligned} x&=0+0z\\ y&=-1+z\\ z&=0+z \end{aligned} $
On
The Maple command $$ solve(\{A_1x+B_1y+C_1z+D_1=0,A_2x+B_2y+C_2z+D_2=0\},\{x,y\}) $$ produces the answer $$ \left\{ x={\frac {{\it B_1}\,{\it C_2}\,z-{\it B_2}\,{\it C_1}\,z+{\it B_1 }\,{\it D_2}-{\it D_1}\,{\it B_2}}{{\it A_1}\,{\it B_2}-{\it A_2}\,{\it B_1}} },y=-{\frac {{\it A_1}\,{\it C_2}\,z-{\it A_2}\,{\it C_1}\,z+{\it A_1}\,{ \it D_2}-{\it A_2}\,{\it D_1}}{{\it A_1}\,{\it B_2}-{\it A_2}\,{\it B_1}}} \right\} .$$ The command $$ solve(\{A_1x+B_1y+C_1z+D_1=0,A_2x+B_2y+C_2z+D_2=0\},\{x,y\},parametric=full) $$ investigates all possibilities depending on the coefficients.
What you look for is a point on the line $\left(x_0, y_0, z_0\right)$, and a direction vector of the line $\left(a, b, c\right)$.
To find a point on the line, you can for example fix $x$ and find $y,z$ (there are some case where this won't work).
To find a direction vector, note that the vector $\left(A_1,B_1,C_1\right)$ is orthogonal to the first plane therefore to the line. Likewise, $\left(A_2,B_2,C_2\right)$ is orthogonal to the line. If you take their vector product you will get a direction vector.
Another way to find a direction vector, is to find another point on the line and subtract one point from the other.