I want to prove analytically that if two non-parallel lines in $\mathbb{R}^n$ are coplanar (i.e. there is a 2D plane that contains both of them), then they must intersect.
I define two lines to be parallel if their direction vectors can be written as non-zero scalar multiples of each other.
My attempt at formalizing this:
Let $a,b,c\in\mathbb{R}^n$ be points and $\mathbf{s},\mathbf{t},\mathbf{u},\mathbf{v}\in\mathbb{R}^n$ be vectors, where for all $k\neq 0$, we have $\mathbf{s}\neq k\mathbf{t}$ and $\mathbf{u}\neq k\mathbf{v}$.
Given a point $p$, we denote by $\mathbf{p}$ its position vector.
Let $A= \{x\in \mathbb{R}^n: \mathbf{x}=\mathbf{a} + \lambda \mathbf{s} + \mu \mathbf{t} \left( \lambda,\mu\in\mathbb{R}\right) \} $ be a plane that contains the point $a$.
Let $B= \{x\in \mathbb{R}^n: \mathbf{x}=\mathbf{b} + \lambda \mathbf{u} \left( \lambda \in\mathbb{R}\right) \} $ and $C= \{x\in \mathbb{R}^n: \mathbf{x}=\mathbf{c} + \lambda \mathbf{v} \left( \lambda\in\mathbb{R}\right) \} $ be non-parallel lines that contain the points $b$ and $c$.
Claim. If $B,C\subseteq A$, then $B\cap C \neq \varnothing$.
(In words: If the plane $A$ contains both lines, then the two lines must intersect.)
How do I prove the above claim? (In particular, what exactly is the intersection point of the two lines in terms of the given variables?)
I omit the boldface notion for convenience.
From the assumptions it follows that with some constants $\alpha_i, \beta_j$ we have
$b=a+\alpha_1 s+\beta_1 t$, $c=a+\alpha_2 s+\beta_2 t$ since $b,c\in A$. Likewise $b+u=a+\alpha_3 s+\beta_3 t$, $c+v=a+\alpha_4 s+\beta_4 t$ since $b+u,c+v\in A$. From this it follows that $u=\alpha_5 s+\beta_5 t$ and $v=\alpha_6 s+\beta_6 t$ again with constants $\alpha_5,\alpha_6,\beta_5,\beta_6$ which can be calculated from the given data.
Looking for the intersection of the lines leads to the task of finding $\lambda,\mu$ such that $b+\lambda u=c+\mu v$. This can be solved uniquely since $s,t$ are linearly independent, and so are $u,v$.