Parallel plane that contain lines

Question

I have the lines:

$$L_1: \frac{x-3}{2}= \frac{y+5}{-3} = \frac{z+1}{5} ,$$ $$L_2: \frac{x+1}{-4}=\frac{y-1}{3}=\frac{z-3}{-1}$$

I need the equations of the parallel planes $P_1$ and $P_2$ that contains those lines. $P_1$ contains $L_1$ and $P_2$ contains $L_2$.

What I did so far:

I got the vector equation of the lines:

$$L_1: (3,-5, -1) + t(2, -3, 5)$$ $$L_2: (-1, 1, 3) + r(-4, 3, -1)$$

The planes are parallel, so $N_1 = N_2 \cdot k, N_1$ and $N_2$ are the normals of both planes.

Because the lines are in the planes, the direction vector of $L_1$ is perpendicular to $N_1$ and the direction vector of $L_2$ is perpendicular to $N_2$. So:

$$N_1 \cdot (2, -3, 5) = 0,$$ $$N_2 \cdot (-4, 3, -1) = 0$$

if $N_1 = N_2 = (x, y, z)$ then $$2x - 3y + 5z = 0$$ $$-4x + 3y -z = 0$$

Here is where I get lost. My idea of a next step was getting the $P_0$ which would be the intersection between $N_1$ and $L_1$ for $P_1$.

I replaced the parametric equations of $L_1$ in one of the last $2$ equations and got that
$$t = \frac{-8}{19}$$ $$P_0 = \left(\frac{41}{19}, \frac{-71}{19}, \frac{21}{19}\right)$$

But I don't know how to get $N_1$. With $P_0$ and $N_1$ I can get the equations.

I didn't try with the next one because I don't think I'm going anywhere with that step.
Thanks for the help.

Answer 1

The two lines are not parallel. Thus you can find the normal to the two desired planes by taking the cross product of direction vectors of the two lines. One you have the normal you just need to substitute a point to find the equation of the plane.

Answer 2

Continuing your work, solving $2x-3y+5z=0$ and $-4x+3y-z=0$

by first adding the equations gives $-2x+4z=0$, so $x=2z$.

Substituting into the 1st equation gives $-3y+9z=0$, so $y=3z$.

If we take $z=1$ for convenience, we get that $N_1=N_2=\langle 2, 3, 1\rangle$.

Now use the points $(3,-5,-1)$ and $(-1,1,3)$ to find equations for the two planes.