The problem asks us to find the equation of a plane that contains two given parallel lines. The parametric equations of the two lines are given as: Line 1: $x = -2 + 2t$ $y = 1 + 4t$ $z = 2 - t$ Line 2: $x = 2 - 2t$ $y = 3 - 4t$ $z = 1 + t$

1. Problem Description

The problem asks us to find the equation of a plane that contains two given parallel lines. The parametric equations of the two lines are given as:

Line 1:

$x = -2 + 2t$

$y = 1 + 4t$

$z = 2 - t$

Line 2:

$x = 2 - 2t$

$y = 3 - 4t$

$z = 1 + t$

2. Solution Steps

To find the equation of the plane, we need a point on the plane and a normal vector to the plane.

First, let's find a point on each line. When $t=0$,

Point on Line 1: $P_1 = (-2, 1, 2)$

Point on Line 2: $P_2 = (2, 3, 1)$

Now we can find a vector connecting these two points:

$\vec{P_1P_2} = P_2 - P_1 = (2 - (-2), 3 - 1, 1 - 2) = (4, 2, -1)$

Next, we need to find the direction vector of the lines. From the parametric equations, we can see that the direction vector is $\vec{v} = (2, 4, -1)$.

Now we have two vectors in the plane, $\vec{P_1P_2}$ and $\vec{v}$. We can find the normal vector to the plane by taking the cross product of these two vectors:

$\vec{n} = \vec{P_1P_2} \times \vec{v} = (4, 2, -1) \times (2, 4, -1) = (2(-1) - (-1)(4), -(4(-1) - (-1)(2)), 4(4) - 2(2)) = (-2 + 4, -(-4 + 2), 16 - 4) = (2, 2, 12)$

We can simplify the normal vector by dividing by 2: $\vec{n} = (1, 1, 6)$.

Now we have a normal vector $\vec{n} = (1, 1, 6)$ and a point $P_1 = (-2, 1, 2)$ on the plane. The equation of the plane is given by:

$n_x(x - x_0) + n_y(y - y_0) + n_z(z - z_0) = 0$

$1(x - (-2)) + 1(y - 1) + 6(z - 2) = 0$

$x + 2 + y - 1 + 6z - 12 = 0$

$x + y + 6z - 11 = 0$

$x + y + 6z = 11$

3. Final Answer

The equation of the plane is $x + y + 6z = 11$.