The problem asks us to find the parametric and symmetric equations of a line that passes through a given point and is parallel to a given vector. We will solve problem 5. The given point is $(4, 5, 6)$ and the given vector is $(3, 2, 1)$.

1. Problem Description

The problem asks us to find the parametric and symmetric equations of a line that passes through a given point and is parallel to a given vector. We will solve problem

5. The given point is $(4, 5, 6)$ and the given vector is $(3, 2, 1)$.

2. Solution Steps

To find the parametric equations of a line, we use the formula:

$x = x_0 + at$

$y = y_0 + bt$

$z = z_0 + ct$

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

In this case, $(x_0, y_0, z_0) = (4, 5, 6)$ and $(a, b, c) = (3, 2, 1)$.

Therefore, the parametric equations are:

$x = 4 + 3t$

$y = 5 + 2t$

$z = 6 + t$

To find the symmetric equations of a line, we solve the parametric equations for $t$ and set them equal to each other:

$t = \frac{x - x_0}{a}$

$t = \frac{y - y_0}{b}$

$t = \frac{z - z_0}{c}$

So,

$\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$

In this case, the symmetric equations are:

$\frac{x - 4}{3} = \frac{y - 5}{2} = \frac{z - 6}{1}$

3. Final Answer

Parametric equations:

$x = 4 + 3t$

$y = 5 + 2t$

$z = 6 + t$

Symmetric equations:

$\frac{x - 4}{3} = \frac{y - 5}{2} = \frac{z - 6}{1}$