The problem asks to find the equation of a line given two points in 3D space. The two points are $A(2, 1, 4)$ and $B(4, 1, -2)$.

1. Problem Description

The problem asks to find the equation of a line given two points in 3D space. The two points are $A(2, 1, 4)$ and $B(4, 1, -2)$.

2. Solution Steps

First, we need to find the direction vector of the line. The direction vector $\vec{v}$ can be found by subtracting the coordinates of point A from the coordinates of point B:

$\vec{v} = \vec{B} - \vec{A} = (4-2, 1-1, -2-4) = (2, 0, -6)$.

The equation of a line in 3D space can be expressed in parametric form as:

$x = x_0 + at$

$y = y_0 + bt$

$z = z_0 + ct$

where $(x_0, y_0, z_0)$ is a point on the line and $(a, b, c)$ is the direction vector.

We can use point A $(2, 1, 4)$ as the point on the line, and the direction vector $\vec{v} = (2, 0, -6)$.

So, the parametric equations of the line are:

$x = 2 + 2t$

$y = 1 + 0t$

$z = 4 - 6t$

This simplifies to:

$x = 2 + 2t$

$y = 1$

$z = 4 - 6t$

3. Final Answer

The parametric equations of the line are:

$x = 2 + 2t$

$y = 1$

$z = 4 - 6t$