We are asked to find the equation of the plane that contains two parallel lines given in parametric form. The equations for the first line are $x = -2 + 2t$, $y = 1 + 4t$, and $z = 2 - t$. The equations for the second line are $x = 2 - 2t$, $y = 3 - 4t$, and $z = 1 + t$.

1. Problem Description

We are asked to find the equation of the plane that contains two parallel lines given in parametric form. The equations for the first line are $x = -2 + 2t$, $y = 1 + 4t$, and $z = 2 - t$. The equations for the second line are $x = 2 - 2t$, $y = 3 - 4t$, and $z = 1 + t$.

2. Solution Steps

First, we find the direction vector of the lines. We can find it by looking at the coefficients of $t$ in the parametric equations. The direction vector $\vec{v}$ is given by $\vec{v} = \langle 2, 4, -1 \rangle$.

Next, we need to find a point on each line. For the first line, when $t=0$, we have the point $P_1 = (-2, 1, 2)$. For the second line, when $t=0$, we have the point $P_2 = (2, 3, 1)$.

Now we can find a vector connecting the two points $P_1$ and $P_2$. Let's call this vector $\vec{w}$.

$\vec{w} = \vec{P_1P_2} = \langle 2 - (-2), 3 - 1, 1 - 2 \rangle = \langle 4, 2, -1 \rangle$.

To find the normal vector $\vec{n}$ to the plane, we take the cross product of the direction vector $\vec{v}$ and the vector $\vec{w}$.

$\vec{n} = \vec{v} \times \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & -1 \\ 4 & 2 & -1 \end{vmatrix} = \hat{i}(4(-1) - (-1)(2)) - \hat{j}(2(-1) - (-1)(4)) + \hat{k}(2(2) - 4(4)) = \hat{i}(-4 + 2) - \hat{j}(-2 + 4) + \hat{k}(4 - 16) = -2\hat{i} - 2\hat{j} - 12\hat{k}$.

Thus, $\vec{n} = \langle -2, -2, -12 \rangle$.

We can simplify this vector by dividing by $-2$, giving us $\vec{n} = \langle 1, 1, 6 \rangle$.

The equation of the plane is given by $ax + by + cz = d$, where $\langle a, b, c \rangle$ is the normal vector to the plane.

So we have $1x + 1y + 6z = d$.

To find $d$, we can use one of the points on the plane. Let's use $P_1 = (-2, 1, 2)$.

$(-2) + (1) + 6(2) = d$, which gives $-2 + 1 + 12 = d$, so $d = 11$.

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

3. Final Answer

$x + y + 6z = 11$