I have two lines:
$$ a: \left\{ \begin{array}{c} x=\lambda+2\mu \\ y=-\lambda+\mu \\ z=0\lambda-\mu \\ t=2\lambda+\mu \end{array} \right. \\ b: \left\{ \begin{array}{c} x=-p-3q \\ y=-p+2q \\ z=p \\ t=4p-q \end{array} \right. $$
How can I find equation of a plane passing those two lines?
In $R^4$
with
$$ p_1 = \{1, -1, 0, 2\}\\ p_2 = \{2, 1, -1, 1\}\\ q_1 = \{-1, -1, 1, 4\}\\ q_2 = \{-3, -2, 0, -1\} $$
we have
$$ L_1\rightarrow p=p_1\lambda+p_2\mu\\ L_2\rightarrow p=q_1\alpha+q_2\beta $$
The plane equation is
$$ P \rightarrow < p - p_0, \vec v > = 0 \ \ \mbox{with} \ \ \vec v = \{v_1,v_2,v_3,v_4\} $$
If the lines are contained into the plane then
$$ < p_1\lambda+p_2\mu-p_0,\vec v > = 0\\ < q_1\alpha+q_2\beta-p_0,\vec v > = 0\\ $$
for all $\lambda,\mu,\alpha,\beta$ and the conditions are
$$ < p_1,\vec v > = 0\\ < p_2, \vec v > = 0\\ < q_1, \vec v > = 0\\ < q_2, \vec v > =0 $$
Solving for $\vec v$ we obtain $\vec v = 0$ hence no plane contains $L_1, L_2$