<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] Note: 
I get 
$$
y + 2z = 0 \Rightarrow z = - y/2 \\
3 + z = (x + y + z) + z = x + y + 2z = x
$$
So 
$$
\frac{x - 3}{1} = \frac{y-0}{-2} = \frac{z - 0}{1}
$$
or
$$
(x, y, z) = (3,0,0) + t (1, -2, 1)
$$
Hint: The projection onto $z=0$ should affect what coordinate?
Spoiler:

 The $z$-coordinate is forced to $0$. So we have $(x,y,z) = (3,0,0) + t (1,-2,0)$. We have $y = - 2 t$ and $x - 3 = t$ so $y = -2(x-3) = -2x + 6$ is the equation of the projected line within the $x$-$y$-plane.

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