I need to find out the relative positions to each other a straight line, first I'm trying to check if they are coplanar but I get an unknown variable.
Can anyone help me on how to solve this part of the resolution?
$r1: x + 3 = \dfrac{2y - 4}4 = \dfrac{z - 1}3 = \lambda$
$r1: x + 3 = \dfrac{y - 2}2 = \dfrac{z - 1}3 = \lambda$
$r1: (x, y, z) = (-3, 2, 1) + \lambda(a, b, c)$
$r2: x = (0, 2, 2) + \lambda(1, 1, -1)$
Checks if are coplanar
$\vec{v_1}\cdot (\vec{v_2} \times \vec{u}) = \begin{vmatrix} a & b & c \\ 1 & 1 & -1 \\ -3 & 0 & 1 \end{vmatrix} = 0?$
Firstly, to correct some things. The first vector equation is: $$r_1: (x,y,z) = \underbrace{(-3,2,1)}_{P_1} +\lambda \cdot \underbrace{(1,2,3)}_{\mathbf a}, \quad \lambda \in \mathbb R.$$
I suppose that the second vector equation is: $$r_2 : (x,y,z) = \underbrace{(0,2,2)}_{P_2} + \lambda \cdot \underbrace{(a,b,c)}_{\mathbf b}, \quad \lambda \in \mathbb R.$$
That is: $$\begin{vmatrix} -3 & 0 & -1 \\ 1 & 2 & 3 \\ a & b & c\end{vmatrix}=0.$$
The above zero determinant implies that all the rows are linearly dependent, thus we have that $(a,b,c)$ can be written as a linear combination of the $2$ other rows, i.e. $$(a,b,c) = \mu \cdot (-3,0,-1) + \nu \cdot (1,2,3),\quad \mu, \nu \in \mathbb R.$$
So, you can consider any $\mu$ and $\nu $ you want (of course it cannot hold $\mu= \nu =0)$, therefore you can define all of $a,b,c$.