The line equation given is $x - z = y = z$. Let's rewrite this equation in a clearer form for easier interpretation:

$x - z = y = z$

We can break it down into two parts:

1. $x - z = y$
2. $y = z$

From equation $y = z$, we substitute $z$ for $y$ in the first equation:

$x - z = z$

Simplifying this, we get:

$x = 2z$

Now we have $x = 2z$ and $y = z$, which gives us parametric equations:

$x = 2t, \quad y = t, \quad z = t$

where $t$ is a parameter.

The direction vector of this line is given by the coefficients of $t$, which are $(2, 1, 1)$. To find the unit vector, we normalize this direction vector.

The magnitude of the vector $\vec{v} = (2, 1, 1)$ is:

$|\vec{v}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$

The unit vector is:

$\hat{v} = \frac{1}{|\vec{v}|} \vec{v} = \frac{1}{\sqrt{6}} (2, 1, 1)$

Thus, the unit vector parallel to the line is:

$\hat{v} = \left( \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right)$

Let me know if you'd like more details or have any questions! Here are some related questions to expand your understanding:

1. What is the geometric interpretation of a unit vector?
2. How would you find the unit vector for a line in 2D space?
3. What is the significance of normalizing a vector?
4. Can you express a unit vector in terms of angles?
5. How would the unit vector change if the direction vector were different?

Tip: To normalize a vector, divide each component of the vector by its magnitude.