Correct Answer: B (0)

The result can be found by using a key property of the cube roots of any number and a common technique for simplifying determinants.

Step 1: Analyze the Cube Roots of 27

The problem states that $x, y,$ and $z$ are the three cube roots of 27. This means they are the solutions to the equation:

$k^3 = 27 \implies k^3 - 27 = 0$

We can rewrite this equation to show the coefficients of the other powers of $k$:

$1k^3 + 0k^2 + 0k - 27 = 0$

According to Vieta's formulas, for any cubic equation $ak^3 + bk^2 + ck + d = 0$, the sum of the roots is equal to $-b/a$.

For our equation, the sum of the roots $x, y,$ and $z$ is:

$x + y + z = -\frac{0}{1} = 0$

This is a crucial property we will use to solve the problem.

Step 2: Simplify the Determinant

The determinant we need to evaluate is:

$D = \begin{vmatrix} x & y & z \\ y & z & x \\ z & x & y \end{vmatrix}$

A standard technique to simplify such determinants is to perform column or row operations. Let's apply the column operation $C_1 \to C_1 + C_2 + C_3$, which means adding the second and third columns to the first column. This does not change the value of the determinant.

$D = \begin{vmatrix} x+y+z & y & z \\ y+z+x & z & x \\ z+x+y & x & y \end{vmatrix}$

Now, we can factor out the common term $(x+y+z)$ from the first column:

$D = (x+y+z) \begin{vmatrix} 1 & y & z \\ 1 & z & x \\ 1 & x & y \end{vmatrix}$

Step 3: Calculate the Final Value

From Step 1, we established that the sum of the roots is $x+y+z = 0$. Substituting this value into our simplified expression for the determinant:

$D = (0) \begin{vmatrix} 1 & y & z \\ 1 & z & x \\ 1 & x & y \end{vmatrix} = 0$

Since one of the factors is 0, the entire determinant is 0.