How to calculate the intersection of two planes? How to calculate the intersection of two planes ?
These are the planes and the result is gonna be a line in $\Bbb R^3$:
$x + 2y + z - 1 = 0$
$2x + 3y - 2z + 2 = 0$

From the coefficients of x, y and z of the general form equations, the first plane has normal vector $\begin{pmatrix}1\\2\\1\end{pmatrix}$ and the second has normal vector $\begin{pmatrix}2\\3\\-2\end{pmatrix}$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $v_1,v_2$ is $v_1\times v_2$, so the direction vector of the line of intersection is $$\begin{pmatrix}1\\2\\1\end{pmatrix}\times \begin{pmatrix}2\\3\\-2\end{pmatrix}=\begin{pmatrix}-7\\4\\-1\end{pmatrix}$$Next, we need to find a particular point on the line. Here, since a line that isn't parallel to any coordinate plane passes through all three, you can check if it is parallel to one by using the directional vector. Since here, the line passes through all three planes, we can try $y=0$(since it passes through the $x-z$ plane) and solve the resulting system of linear equations:$$\begin{align}x+z-1&=&0\\2x-2z+2&=&0\end{align}$$ giving $x=0, z=1$, thus the line of intersection is $\lbrace{\begin{pmatrix}-7t\\4t\\1-t\end{pmatrix}:t\in \Bbb R\rbrace}$
