If A = pmatrix y & z & x z & x & y x & y & z pmatrix where x , y ,...

For an orthogonal matrix $A$, the condition $A A^T = I$ must be satisfied.

Given $A = \begin{pmatrix} y & z & x \\ z & x & y \\ x & y & z \end{pmatrix}$, we find $A^T = \begin{pmatrix} y & z & x \\ z & x & y \\ x & y & z \end{pmatrix}$.

Multiplying $A$ and $A^T$ gives:

$A A^T = \begin{pmatrix} y & z & x \\ z & x & y \\ x & y & z \end{pmatrix} \begin{pmatrix} y & z & x \\ z & x & y \\ x & y & z \end{pmatrix} = \begin{pmatrix} x^2+y^2+z^2 & xy+yz+zx & xy+yz+zx \\ xy+yz+zx & x^2+y^2+z^2 & xy+yz+zx \\ xy+yz+zx & xy+yz+zx & x^2+y^2+z^2 \end{pmatrix}$

Equating $A A^T$ to the identity matrix $I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$, we compare the corresponding elements.

From the diagonal elements, we get:

$x^2 + y^2 + z^2 = 1$

From the non-diagonal elements, we get:

$xy + yz + zx = 0$

Since $x$, $y$, and $z$ are integers, the condition $x^2 + y^2 + z^2 = 1$ implies that one of the variables must be $\pm 1$ and the other two must be $0$. This also satisfies $xy + yz + zx = 0$.

Therefore, the value of $x^2 + y^2 + z^2$ is $1$.

Answer: $1$