Solve this

We have, $A=\frac{1}{3}\left[\begin{array}{ccc}1 & 1 & 2 \\ 2 & 1 & -2 \\ x & 2 & y\end{array}\right]$

$\Rightarrow A^{T}=\frac{1}{3}\left[\begin{array}{ccc}1 & 2 & x \\ 1 & 1 & 2 \\ 2 & -2 & y\end{array}\right]$

Now, $A^{T} A=I$

$\Rightarrow\left[\begin{array}{ccc}x^{2}+5 & 2 x+3 & x y-2 \\ 3+2 x & 6 & 2 y \\ x y-6 & 2 y & y^{2}+8\end{array}\right]=\left[\begin{array}{ccc}9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9\end{array}\right]$

The corresponding elements of two equal matrices are not equal. Thus, the matrix $A$ is not orhtogonal.