Given planes:
$$\pi_1: 3x+12y-3z-5=0$$ $$\pi_2: 3x-4y+9z+7=0$$
and lines $$l_1: x=3-2t, y=-1+3t, z=2+3t$$ $$l_2: x=-5+2t, y=3-4t, z=-1+3t$$ find a line that is parallel to the planes and intersects with these two lines.
For the new line to be parallel to two planes its direction vector has to be cross product of $[-2,3,3]$ and $[2,-4,3]$. But then I can choose only one point for the parametric form of this line. But lines $l_1$ and $l_2$ do not intersect. What should I do? Thanks for your help.
The required line is in the direction of $(3,12,-3)\times (3,-4,9)=12(8,-3,-4)$
Suppose that line meets $l_1$ at $(3-2s,-1+3s,2+3s)$ and $l_2$ at $(-5+2t,3-4t,-1+3t)$. Then
$$(3-2s+5-2t,-1+3s-3+4t,2+3s+1-3t)\parallel (8,-3,-4)$$
$$\frac{8-2s-2t}{8}=\frac{-4+3s+4t}{-3}=\frac{3+3s-3t}{-4}$$