Analytic geometry in the space equations

Question

I'm studying now the part of the algebra which talks about perpendicularity and // of the line to the plane. And I am stuck in this. I have the vector $P(1,2,3)$ perpendicular to $2x-3y+z+1=0$. (find the equations of the straight called r, which passes from $P$ and perpendicular to the plain) $P=a(x-1)+b(y-2)+c(z-3)$ and the parameters directors are $l=2,\: m=-3, \:n=1$. So at the end I've got $x-1/2=y-2/-3=z-3/1$. And to this point it's all ok. Buy I can't understand how from the final equation that I wrote I must pass to \begin{cases} x=2z-5 \\ y=-3+11\end{cases} What operations must I do or study?

Answer 1

It is simply some oerations:

$\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z-3}{1}$

$\frac{x-1}{2}=\frac{z-3}{1}$

$x-1=2z-6$ ⇒ $x=2z-5$

$\frac{y-2}{-3}=\frac{z-3}{1}$

$y-2=-3z+9$ ⇒ $y=-3z+11$