The given parametric equations of the line are:

$x = 4 + 7t, \quad y = 5 - 3t, \quad z = 2t$

A vector parallel to the line can be obtained by looking at the coefficients of $t$ in these parametric equations, because the direction of the line is determined by these coefficients.

For $x$, $y$, and $z$, the coefficients of $t$ are $7$, $-3$, and $2$ respectively.

Therefore, the direction vector (which is also parallel to the line) is:

$\mathbf{v} = \langle 7, -3, 2 \rangle$

This is the position vector parallel to the given line.

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Would you like further details or have any questions?

Here are 5 related questions:

1. How do you find the parametric equations of a line given two points?
2. What is the dot product of two vectors and how is it used to find if they are parallel?
3. Can you find a unit vector parallel to the vector $\langle 7, -3, 2 \rangle$?
4. How would you determine if two vectors are perpendicular?
5. What is the scalar projection of one vector onto another?

Tip: When finding a vector parallel to a line, always focus on the coefficients of the parameter from the parametric form of the line.