A problem with planes, normal equations, lines and points. I was given the following problem and would like to know if my idea oh how to solve it is correct. I usually type my answers but due to my sketch I'll attach a picture with my interpretation of the problem and its possible solution 

Question: 
Find the normal of the lines $l$ that pass through a point $(1,0,2)$ and is orthogonal to a plane whose normal equation is $x+y-2z = 4$.



Basically my idea was the following, if I have a line orthogonal to a plane in $R^3$ then the direction vector of such plane will be parallel to my line. Then if I want to find the normal of the lines that pass through a point $P_o$ A line can be held with two vectors $(A,B,C)$ and $(a,b,c)$ whose dot product with the direction vector $(1,1,-2)$ is normal to the line and passes through $P_o$ 

Please, let me know if my work is correct or if I went wrong where did I fail to interpret the problem, what is wrong with my sketch. I genuinely appreciate any help thank you.
$(1,1,-2)$ is normal to your plane.  This is the direction vector of your line.

Vector equation: $L: (1,0,2) + (1,1,-2)t$

Parametric equations: $x = 1+t, y = 0+t, z = 2-2t$

Simultaneous equations: $\frac {x-1}{1} = \frac {y-0}{1}= \frac {z-2}{-2}$
$(1,1,-2)$ is normal to your plane.  This is the direction vector of your line.

Vector equation: $L: (1,0,2) + (1,1,-2)t$

Parametric equations: $x = 1+t, y = 0+t, z = 2-2t$

Simultaneous equations: $\frac {x-1}{1} = \frac {y-0}{1}= \frac {z-2}{-2}$