The values of a, b, c if $\left[\begin{array}{ccc}0& 2b& c\\ a& b& -c\\ a& -b& c\end{array}\right]$ is orthogonal are

$\text{ Let }A=\left[\begin{array}{ccc}0& 2b& c\\ a& b& -c\\ a& -b& c\end{array}\right]\text{. Now, }{A}^T=\left[\begin{array}{ccc}0& a& a\\ 2b& b& -b\\ c& -c& c\end{array}\right]$

 A is orthogonal
$\therefore {\mathrm{AA}}^{\mathrm{T}}=\mathrm{I}\Rightarrow \left[\begin{array}{ccc}0& 2\mathrm{b}& \mathrm{c}\\ \mathrm{a}& \mathrm{b}& -\mathrm{c}\\ \mathrm{a}& -\mathrm{b}& \mathrm{c}\end{array}\right]\left[\begin{array}{ccc}0& \mathrm{a}& \mathrm{a}\\ 2\mathrm{b}& \mathrm{b}& -\mathrm{b}\\ \mathrm{c}& -\mathrm{c}& \mathrm{c}\end{array}\right]=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]$

Equating the corresponding elements, we get 

$\begin{array}{l}4b^2+c^2=1-----\left(1\right)\\ 2b^2-c^2=0----\left(2\right)\\ a^2+b^2+c^2=1----\left(3\right)\end{array}$

On solving (1), (2) and (3) , we get 

$\mathrm{a}=\pm \frac{1}{\sqrt{2}};\mathrm{b}=\pm \frac{1}{\sqrt{6}};\mathrm{c}=\pm \frac{1}{\sqrt{3}}$