Find the equations of the line of intersection of the following planes Find the equations of the line of intersection of the following planes
$2x − 3y + 2z = 5$ and $x + 2y − z = 4$.
So i first put this in the normal vector form
$\langle 2, -3, 2\rangle$
$\langle 1, 2, -1\rangle$
Then i took the cross product which i got
$\langle -1, 4, 7\rangle$
I then made $z = 0$ to solve for $x$ and $y$. Then i subtracted the original equations
$$\begin{matrix}&2x - 3y = 5 \\ - &(x + 2y = 4)\end{matrix}$$
and got
$x-y = 1$
$x = 1+y$
$y = 1-x$
Subbing $x$ and $y$ into the  equation of $2x-3y=5$ i got $x= \frac 85$ and $y = -3$
So far i feel as if this is wrong i was trying to solve this question following a youtube video https://www.youtube.com/watch?v=LpardiBTAvU but if this process is correct i do not know how to proceed.
 A: you almost solved it. you have a mistake with the point on the line. equating z to zero and solving a 2x2 linear system will yield the point on the line ${\bf{P_0}}=(\frac{22}{7},\frac{3}{7}, 0)^T$. Hence, the line of intersection is given by its parametric representation as
\begin{equation}
{\bf{l}}=(\frac{22}{7},\frac{3}{7}, 0)^T + \alpha (−1,4,7)^T \\ 
\forall \alpha \in \mathbb{R}
\end{equation}
A: we consider
$$2x-3y+2z=5$$
$$x+2y-z=4$$
multiplying the second equation by $$-2$$ and adding to the first one we obtain
$-7y+4z=-3$
$$y=\frac{4}{7}z+\frac{3}{7}$$
can you apply this?
A: solving the system of the equations of the two plane we find the common straight line ( if it exists).
$$
\begin{cases}
2x-3y+2z=5\\
x+2y-z=4
\end{cases}
\Rightarrow
\begin{cases}
x=t \;(\forall t \in \mathbb{R})\\
y=13-4x=13-4t\\
z=22-7x=22-7t
\end{cases}
$$
So the straight line has equation:
$$
\begin{bmatrix}
x\\y\\z
\end{bmatrix}
=
\begin{bmatrix}
1\\-4\\-7
\end{bmatrix}
t+
\begin{bmatrix}
0\\13\\22
\end{bmatrix}
$$
