To show that the lines $L_1$ and $L_2$ are parallel and find the distance between them, we'll follow these steps:

Step 1: Represent the Lines in Parametric Form

The given parametric equations are:

• Line $L_1$: $x = 2 - t$, $y = 2t$, $z = 1 + t$
• Line $L_2$: $x = 1 + 2t$, $y = 3 - 4t$, $z = 5 - 2t$

Step 2: Identify the Direction Vectors

For line $L_1$, the direction vector $\mathbf{v}_1$ can be found from the coefficients of $t$: $\mathbf{v}_1 = (-1, 2, 1)$ For line $L_2$, the direction vector $\mathbf{v}_2$ is: $\mathbf{v}_2 = (2, -4, -2)$

Step 3: Check if the Vectors are Parallel

Two vectors are parallel if one is a scalar multiple of the other, i.e., $\mathbf{v}_1 = k \mathbf{v}_2$ for some scalar $k$.

Let's check if $\mathbf{v}_1$ is a scalar multiple of $\mathbf{v}_2$: $(-1, 2, 1) = k(2, -4, -2)$ Comparing each component:

1. $-1 = 2k$ implies $k = -\frac{1}{2}$
2. $2 = -4k$ also gives $k = -\frac{1}{2}$
3. $1 = -2k$ again gives $k = -\frac{1}{2}$

Since all components give the same value for $k$, the vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ are indeed parallel. Thus, the lines $L_1$ and $L_2$ are parallel.

Step 4: Find the Distance Between the Lines

To find the distance $D$ between two parallel lines in space, use the formula: $D = \frac{| (\mathbf{P_1} - \mathbf{P_2}) \cdot (\mathbf{v}_1 \times \mathbf{v}_2) |}{|\mathbf{v}_1 \times \mathbf{v}_2|}$ where:

• $\mathbf{P_1} = (2, 0, 1)$ is a point on line $L_1$
• $\mathbf{P_2} = (1, 3, 5)$ is a point on line $L_2$
• $\mathbf{v}_1 \times \mathbf{v}_2$ is the cross product of $\mathbf{v}_1$ and $\mathbf{v}_2$

Step 4.1: Compute $\mathbf{P_1} - \mathbf{P_2}$

$\mathbf{P_1} - \mathbf{P_2} = (2 - 1, 0 - 3, 1 - 5) = (1, -3, -4)$

Step 4.2: Compute $\mathbf{v}_1 \times \mathbf{v}_2$

The cross product $\mathbf{v}_1 \times \mathbf{v}_2$ is calculated as:

\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ 2 & -4 & -2 \\ \end{vmatrix} = \hat{i}(2(-2) - 1(-4)) - \hat{j}((-1)(-2) - 1(2)) + \hat{k}((-1)(-4) - 2(2))$$ $$= \hat{i}(-4 + 4) - \hat{j}(2 - 2) + \hat{k}(4 - 4) = (0, 0, 0)$$ Since the cross product results in the zero vector, the lines are not only parallel but also lie in the same plane. Hence, the distance between them is zero.