The equation given in the image is:

$x^2 + 13y^2 + 2z^2 - 6xy - 4yz - 6z + 9 = 0.$

To find all the values of $x$, $y$, and $z$, we can attempt various strategies including trial solutions or methods like completing the square. First, let's rearrange and attempt simplifications.

The equation is quadratic in terms of $x$, $y$, and $z$, involving cross terms such as $xy$ and $yz$, and linear terms like $6z$. I will attempt to factor or find a structured way to solve this step-by-step. Let me proceed with the algebra and calculations.

The solutions for the variables $x$, $y$, and $z$ involve the following two expressions for $x$, with $y$ and $z$ being arbitrary:

$x = 3y \pm \sqrt{-4y^2 + 4yz - 2z^2 + 6z - 9}$

This means $y$ and $z$ can be any values, and $x$ depends on both through the above relationship. To explore specific values, you'd need to assign values to $y$ and $z$, and then determine $x$.

Would you like to explore specific values for $y$ and $z$, or do you have any other questions?

Here are some follow-up questions:

1. What values of $y$ and $z$ can simplify the expression for $x$?
2. Can the discriminant of the square root be non-negative for real solutions?
3. How does changing the values of $z$ affect $x$?
4. Can you derive constraints on $y$ and $z$ for real solutions?
5. What happens when $y = z = 0$?

Tip: When dealing with quadratic expressions involving multiple variables, try to simplify cross terms by substituting known values to gain insights.