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 $\Bbb R^3$:
$x + 2y + z - 1 = 0$
$2x + 3y - 2z + 2 = 0$

Let $x$ be in the intersection of the planes
\begin{equation*}
  \langle m, x-b \rangle = 0,\\
  \langle n, x-c \rangle = 0
\end{equation*}
and let $p = m \times n$.
Then for some $\lambda$,
\begin{equation*}
  \begin{pmatrix}
    m_0 & m_1 & m_2 \\
    n_0 & n_2 & n_3 \\
    p_0 & p_1 & p_2 \\
  \end{pmatrix}
  \begin{pmatrix}
    x_0 \\ x_1 \\ x_2
  \end{pmatrix}
  =
  \begin{pmatrix}
    \langle m, b \rangle \\
    \langle n, c \rangle \\
    \lambda
  \end{pmatrix}.
\end{equation*}
Using the vector identities
\begin{equation*}
  \begin{split}
    u \times (v \times w) &= \langle u, w \rangle v - \langle u, v \rangle w, \\
    \Vert u \times v \Vert^2 &= \Vert u \Vert^2 \Vert v \Vert^2 - \langle u, v \rangle^2,
  \end{split}
\end{equation*}
check that
\begin{equation*}
  \begin{pmatrix}
    m_0 & m_1 & m_2 \\
    n_0 & n_1 & n_2 \\
    p_0 & p_1 & p_2 \\
  \end{pmatrix}
  \begin{pmatrix}
    q_0 & r_0 & p_0 \\
    q_1 & r_1 & p_1 \\
    q_2 & r_2 & p_2 \\
  \end{pmatrix}
  = \Vert p \Vert^2 I,
\end{equation*}
where
\begin{equation*}
  \begin{split}
    q &= -n \times (n \times m), \\
    r &= -m \times (m \times n).
  \end{split}
\end{equation*}
If $p \ne 0$ (that is, if $m$ and $n$ are linearly independent), it follows that
\begin{equation*}
  \begin{pmatrix}
    x_0 \\ x_1 \\ x_2
  \end{pmatrix}
  = \frac{1}{\Vert p \Vert^2}
  \begin{pmatrix}
    q_0 & r_0 & p_0 \\
    q_1 & r_1 & p_1 \\
    q_2 & r_2 & p_2 \\
  \end{pmatrix}
  \begin{pmatrix}
    \langle m, b \rangle \\
    \langle n, c \rangle \\
    \lambda
  \end{pmatrix},
\end{equation*}
or
\begin{equation*}
  x = a + \lambda \frac{p}{\Vert p \Vert^2}
\end{equation*}
where
\begin{equation*}
  a = (\langle m, b \rangle q + \langle n, c \rangle r) / {\Vert p \Vert^2}.
\end{equation*}
