If the matrix $A=\begin{bmatrix} 0 & 2y & z \\x & y & -z \\ x & -y & z \end{bmatrix}$ satisfies the equation $A^TA=I_3,$ then $x^2+y^2 +z^2 $ is :

1

The correct answer is Option (2) → 1

$AA^T=I_3$    [Given]

$\begin{bmatrix} 0 & x & x \\2y & y & -y \\ z & -z & z \end{bmatrix}\begin{bmatrix} 0 & 2y & z \\x & y & -z\\ x & -y & z \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$

$\begin{bmatrix} 2x^2 & 0 & 0 \\0 & y^2 & 0 \\ 0 & 0 & 3z^2 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$

$x^2=\frac{1}{2},y^2=\frac{1}{6},z^2=\frac{1}{3}$

$∴x^2+y^2+z^2=\frac{1}{2}+\frac{1}{6}+\frac{1}{3}=1$