How to prove that 3D lines given by their equations are noncoplanar $(R^3,(R^3,q_0),φ)$ 
picture for the $(R^3,(R^3,q_0),φ)$  

$D_1: \ x_1-3=x_2/2=(x_3-1)/3 $

$D_2: \ x_1=x_2/2=x_3$  

Q: Is it true $D_1,D_2$ are noncoplanar ? If yes, why ?

Hi! Has anybody an idea about a formula I should use or an advice on how to proceed ?
First note that their direction vectors are different. It follows that they cannot be parallel. Then note that they never intersect.

The two facts above imply they are skew. Thus, they cannot be coplanar.
They cannot intersect because there cannot exist a common point $(x_1,x_2,x_3)$ to them ; indeed, it should verify the different relationships.

But the two first equalities are already incompatible : it is not possible that coordinates $(x_1,x_2)$ verify simultaneously :

$$x_1-3=x_2/2 \ \ \text{and} \ \ x_1=x_2/2\tag{1}$$

It would mean that $x_1-3=x_1$...

(Please note that we haven't used $x_3$ for obtaining the impossibility of a common point).