Answer: \((a,b,c)=(2,-2,1)\) or \((-2,2,-1)\); \(A^{-1}=A^T\).
For an orthogonal matrix the columns are unit vectors and mutually perpendicular (\(A^TA=I\)).
Columns 1 and 2: \(C_1=\frac13(1,2,2)\), \(C_2=\frac13(2,1,-2)\); \(C_1\cdot C_2=\frac19(2+2-4)=0\) ✓ and each has length 1 ✓.
\(C_3=\frac13(a,b,c)\) must satisfy
\(C_1\cdot C_3=0:\ a+2b+2c=0\) ...(1); \(C_2\cdot C_3=0:\ 2a+b-2c=0\) ...(2); \(|C_3|=1:\ a^2+b^2+c^2=9\) ...(3).
By cross-multiplication from (1) and (2): \(\dfrac{a}{2(-2)-2(1)}=\dfrac{b}{2(2)-1(-2)}=\dfrac{c}{1(1)-2(2)}\Rightarrow\dfrac a{-6}=\dfrac b6=\dfrac c{-3}=k\), i.e. \(a=-6k,\ b=6k,\ c=-3k\).
(3): \(36k^2+36k^2+9k^2=81k^2=9\Rightarrow k=\pm\frac13\). So \((a,b,c)=(2,-2,1)\) or \((-2,2,-1)\).
Taking \((2,-2,1)\): \(A=\frac13\begin{bmatrix}1&2&2\\2&1&-2\\2&-2&1\end{bmatrix}\), and \(A^{-1}=A^T=\frac13\begin{bmatrix}1&2&2\\2&1&-2\\2&-2&1\end{bmatrix}\) (this \(A\) happens to be symmetric). For \((-2,2,-1)\): \(A^{-1}=A^T=\frac13\begin{bmatrix}1&2&2\\2&1&-2\\-2&2&-1\end{bmatrix}\).
