Find the symmetric equations and the parametric equations of a line, and/or the equation of the plane satisfying the following given conditions. Line through (5,-4,2) and parallel to the line \(\mathbf{r}=\mathbf{i}-\mathbf{j}+(5 \mathbf{i}-2 \mathbf{j}+\mathbf{k}) t\).

Symmetric equations are a way to represent a line in three-dimensional space. They are derived from parametric equations by eliminating the parameter, usually represented by \(t\). For symmetric equations, we begin by solving each of the parametric equations for the parameter \(t\) and then equating them to form a single expression.

For example, given the parametric equations:
\(x = 5 + 5t\)
\(y = -4 - 2t\)
\(z = 2 + t\)
We solve each for \(t\):
\(t = \frac{x - 5}{5}\)
\(t = \frac{y + 4}{-2}\)
\(t = z - 2\)
Setting these equations equal to each other gives the symmetric equations:
\[\frac{x - 5}{5} = \frac{y + 4}{-2} = z - 2\]
Simplifying these equations to find the parameter \(t\) ensures that we accurately describe the line through a common equation.

The direction vector of a line indicates the direction the line is pointing in three-dimensional space. It is a vector that, when added to a point on the line, gives another point on the line by scaling it with a parameter, typically \(t\).

Given a line described parametrically as:
\(\textbf{r} = \textbf{a} + \textbf{d}t\)
Where \(\textbf{a}\) is a point on the line and \(\textbf{d}\) is the direction vector. The direction vector \(\textbf{d}\) is essentially the vector formed by the coefficients of \(t\) in the parametric equation.

In the problem, the given line is:
\(\textbf{r} = (1, -1, 0) + t(5, -2, 1)\)
Hence, the direction vector is \((5, -2, 1)\). For the new line through the point \((5, -4, 2)\) and parallel to the given line, the direction vector remains \((5, -2, 1)\).

Parametric equations of a line express the coordinates of points on the line as functions of a single parameter, usually \(t\). They show how the coordinates change with respect to the parameter.

For a line passing through a point \((x_0, y_0, z_0)\) with a direction vector \((a, b, c)\), the parametric equations can be written as:
\(x = x_0 + at\)
\(y = y_0 + bt\)
\(z = z_0 + ct\)
In our problem, the line passes through the point \((5, -4, 2)\) with direction vector \((5, -2, 1)\). Thus, the parametric equations are:
\(x = 5 + 5t\)
\(y = -4 - 2t\)
\(z = 2 + t\)
These equations show the position of any point on the line as \(t\) varies.