Find the line by a given point $(1, 2, 3)$, the line has common points with $\frac{x}{2}=\frac{y+1}{-2}=\frac{z-2}{1}$ and $\frac{x}{4}=\frac{y+2}{0}=\frac{z}{3}$
What am I doing wrong?
I decide to give a line as an intersection of two planes. A plane can be defined by a line and a point that is not on that line.
Consider a plane defined by $(1, 2, 3)$ and $\frac{x}{2}=\frac{y+1}{-2}=\frac{z-2}{1}$. If $\frac{x}{2}=\frac{y+1}{-2}=\frac{z-2}{1}$ lies in the plane, then the plane's normal vector can be any vector that is orthogonal to $\{2, -2, 1\}$; for example, $\{2, 1, -2\}=\{A, B, C\}$. Using $Ax+By+Cz+D=0$, we get $2x+y-2z+2=0$.
Consider a plane defined by $(1, 2, 3)$ and $\frac{x}{4}=\frac{y+2}{0}=\frac{z}{3}$. Analogically, we get $3x+y-4z+7=0$.
The answer in the book is $\begin{cases} 5x+y-8z+17=0 \\ 12x+9y-16z+18=0 \end{cases}$ , which didn't confuse me (an infinite number of intersecting planes can give a single line) until I saw that the direction vectors of my answer and the book's answer don't match. In fact, the direction vector of my answer is $\{2, -2, 1\}$, which is just the first line from the exercise.
I suspect I went astray with the second line, but maybe the entire reasoning is wrong.
Thank you.
This is not true. Since the line is in the plane the normal vector is orthogonal to $(2,-2,1).$ But it can't be any orthogonal vector. Since $(1,2,3)$ and $(0,-1,2)$ are points of the plane the normal vector is also perpendicular to $(1,3,1).$ So, the normal vector is
$$(2,-2,1)\times (1,3,1)=(-5,-1,8).$$