<s>[INST] Find the projection of the line $x+y+z-3=0=2x+3y+4z-6$ on the plane $z=0$ 
Find the projection of the line $x+y+z-3=0=2x+3y+4z-6$ on the plane $z=0$

The equation represents the line of intersection of two planes. Using augmented matrix
$$
\begin{bmatrix} 
1 & 1 & 1 & 3 \\
2 & 3 & 4 & 6
\end{bmatrix}$$
$R_2\rightarrow R_2-2R_1$
$$
\begin{bmatrix} 
1 & 1 & 1 & 3 \\
0 & 1 & 2 & 0
\end{bmatrix}$$
$$y+2z=0$$
$$x+y+z=3$$
Equation of line is
$$\frac{x-0}{1}=\frac{y-0}{-2}=\frac{z-0}{1}$$
Angle made by line and normal is $$\cos(90-\theta)=\frac{1}{\sqrt{6}}$$
where $\theta$ is the angle made by line and plane.
How should I proceed?
 [/INST] A line passes through two points.
You can get the first point intersecting your line with the plane $z=0$. $p=(3,0,0)$.
Simply project another random point of your line, say $(0,6,-3)$, in the plane $z=0$, you get $q=(0,6,0)$.
The line you are looking for is thus $z=0$, $y=-2x+6$.
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