Intersection of two dimensional lines in 3d and create a plane with conditions. 
I'm asked to find the equation of plane satisfying the given conditions:
  
  
*
  
*Passing through the line given by:
  \begin{cases}
x+y=2 \\
y-z=3
\end{cases}
  
*Perpendicular to the plane:
  $$
2 x+3 y+4 z=5
$$
  Knowing that the normal to the plane is 
  $2 i+3 j+4 k$
  

I would have hade no problems finding this out if I was given the point. However I am not able to figure it out.
My first tought was to find the point where these lines intersect and then use this point to create the plane with these coinditions, 
$$
x+y-2=y-z-3\Rightarrow z=-x-1
$$
Which I could have expected since I am dealing with tree variables. 
Now how could I solve this?
Answer should be $x+6 y-5 z=17$
 A: In 3d an equation in the form $Ax+By+Cz=D$ specifies a plane where $A^2+B^2+C^2\neq 0$. Vector $\mathbf{n}=A\mathbf{i}+B\mathbf{j}+C\mathbf{k}$ is called "normal vector" to that plane 'cause it's perpendicular to every vector that lies in the plane. Short form of $Ax+By+Cz=D$ is then $\mathbf{n}\cdot(\mathbf{x}-\mathbf{a})=0$ where $\mathbf{x}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$ and $\mathbf{a}$ is position vector to an arbitrary point of the plane.
A line can be specified in at least 3 ways:


*Two planes intersection: $$\begin{cases}a_1x+b_1y+c_1z=d_1\\ a_2x+b_2y+c_2z=d_2\end{cases},$$

*Parametric form: $$\begin{cases}x=tl_x+a_x\\y=tl_y+a_y\\z=tl_z+a_z\end{cases},$$ or which is equivalent to $$\mathbf{x}=t\mathbf{l}+\mathbf{a},$$ where $\mathbf{l}$ is "directing vector" and $\mathbf{a}$ is selected point on the line and $\mathbf{x}$ is an arbitrary point on the line and $t$ is a number("the parameter").

*"Canonical" form: $$\frac{x-a_x}{l_x}=\frac{y-a_y}{l_y}=\frac{z-a_z}{l_z}(=t)$$
The latter 2 forms are almost equivalent. One should be certain how to derive any of them from "two planes intersection":


*

*Find directing vector of the line as it's perpendicular to both normals of the planes, as their cross product $\mathbf{l}=[\mathbf{n}_1\times \mathbf{n}_2]$

*Find a point \mathbf{a} on the line i.e. on either of those two planes. It could be done letting one of the coordinates to be $0$ or any other way.

*Construct the equation: $\mathbf{x}=\mathbf{a}+t\mathbf{l}$


So we've found the line equation $$\frac{x-2}{1}=\frac{y-0}{-1}=\frac{z+3}{-1}=t\hbox{.}  $$
Then we know that the desired plane's normal vector is perpendicular to the line directing vector ('cause it's perpendicular to every vector within itself) and is perpendicular to the given plane's normal vector $2\mathbf{i}+3\mathbf{j}+4\mathbf{k}$('cause the desired plane and the given plane are given to be perpendicular), so it must be parallel to their cross product: $\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\1&-1&-1\\2&3&4\end{array}\right|=-\mathbf{i}-6\mathbf{j}+5\mathbf{k}$
One thing is missing to write down $\mathbf{n}\cdot(\mathbf{x}-\mathbf{a})=0$: the point $\mathbf{a}$.
Note that the intersection point of the given line and plane also lays on the desired plane 'cause the entire given line does. Let's find it:
$$\begin{cases}\mathbf{x}=2\mathbf{i}+0\mathbf{j}-3\mathbf{k}+t(\mathbf{i}-\mathbf{j}-\mathbf{k})\\(2\mathbf{i}+3\mathbf{j}+4\mathbf{k})\cdot \mathbf{x} = 5\end{cases}$$
$$\left(2\mathbf{i}+0\mathbf{j}-3\mathbf{k}+t(\mathbf{i}-\mathbf{j}-\mathbf{k})\right)\cdot(2\mathbf{i}+3\mathbf{j}+4\mathbf{k}) = 5  $$
$$4-12+t(2-3-4)=5$$
$$t=\frac{13}{5}$$
$$\mathbf{a}=2\mathbf{i}+0\mathbf{j}-3\mathbf{k}+\frac{13}{5}(\mathbf{i}-\mathbf{j}-\mathbf{k})=\frac{23}{5}\mathbf{i}-\frac{13}{5}\mathbf{j}-\frac{28}{5}\mathbf{k},  $$
$$\mathbf{n}\cdot\mathbf{a}=(-\mathbf{i}-6\mathbf{j}+5\mathbf{k})\cdot\left(\frac{23}{5}\mathbf{i}-\frac{13}{5}\mathbf{j}-\frac{28}{5}\mathbf{k}\right)=-\frac{85}{5}=-17  $$
Which gives the desired plane's equation $\mathbf{n}\cdot\mathbf{x}=\mathbf{n}\cdot\mathbf{a}: -x-6y+5x=-17$.  
Questions welcome as any edits for my improper math English do either.
