$A=\left[\begin{array}{ccc}0 & 2 y & z \\ x & y & -z \\ x & -y & z\end{array}\right]$

$A^{\prime}=\left[\begin{array}{ccc}2 y & y & -y \\ z & -z & z\end{array}\right]$

$I=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\mathrm{Now}, \mathrm{A}^{\prime} \mathrm{A}=\mathrm{I}$

Putting values

$\left[\begin{array}{ccc}0 & x & x \\2 y & y & -y \\z & -z & z\end{array}\right]\left[\begin{array}{ccc}0 & 2 y & z \\x & y & -z \\x & -y & z\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]$

$\left[\begin{array}{ccc}0(0)+x(x)+x(x) & 0(2 y)+x(y)+x(-y) & 0(z)+x(-z)+x(z) \\ 2 y(0)+y(x)-y(x) & 2 y(2 y)+y(y)-y(-y) & 2 y(z)+y(-z)-y(z) \\ z(0)-z(x)+z(x) & z(2 y)-z(y)+z(-y) & z(z)-z(-z)+z(z)\end{array}\right]=\left[\begin{array}{ccccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right.$

$\left[\begin{array}{ccc}0+x^{2}+x^{2} & 0+x y-x y & 0-x z+x z \\ 0+x y-x y & 4 y^{2}+y^{2}+y^{2} & 2 z y-z y-z y \\ 0-x z+x z & 2 z y-z y-z y & z^{2}+z^{2}+z^{2}\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\left[\begin{array}{ccc}2 x^{2} & 0 & 0 \\0 & 6 y^{2} & 0 \\0 & 0 & 3 z^{2}\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]$

since matrices are equal,

corresponding elements are equal

$\begin{array}{l|l|l}2 x^{2}=1 & 6 y^{2}=1 & 3 z^{2}=1 \\x^{2}=\frac{1}{2} & y^{2}=\frac{1}{6} & z^{2}=\frac{1}{3} \\x=\pm \sqrt{\frac{1}{2}} & y=\pm \sqrt{\frac{1}{6}} & z=\pm \sqrt{\frac{1}{3}} \\x=\pm \frac{1}{\sqrt{2}} & y=\pm \frac{1}{\sqrt{6}} & z=\pm \frac{1}{\sqrt{3}}\end{array}$

Thus, $x=\pm \frac{1}{\sqrt{2}}, y=\pm \frac{1}{\sqrt{6}}, z=\pm \frac{1}{\sqrt{3}}$