If $\left[\begin{array}{ccc}0& 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{array}\right]$ is orthogonal, then the value of $2{\alpha }^2 + 6{\beta }^2 + 3{\gamma }^2$

Let $A = \left[\begin{array}{ccc}0& 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{array}\right],$then $A\text{'} = \left[\begin{array}{ccc}0& \alpha & \alpha \\ 2\beta & \beta & -\beta \\ \gamma & -\gamma & \gamma \end{array}\right]$

Since, $A$is orthogonal. 

$$\begin{gathered} \therefore AA\text{'} = I \\ \Rightarrow \left[\begin{array}{ccc}0& 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{array}\right] \left[\begin{array}{ccc}0& \alpha & \alpha \\ 2\beta & \beta & -\beta \\ \gamma & -\gamma & \gamma \end{array}\right] = \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \\ \Rightarrow \left[\begin{array}{ccc}4{\beta }^2+{\gamma }^2& 2{\beta }^2-{\gamma }^2& -2{\beta }^2+{\gamma }^2\\ 2{\beta }^2-{\gamma }^2& {\alpha }^2+{\beta }^2+{\gamma }^2& {\alpha }^2-{\beta }^2-{\gamma }^2\\ -2{\beta }^2+{\gamma }^2& {\alpha }^2-{\beta }^2-{\gamma }^2& {\alpha }^2+{\beta }^2+{\gamma }^2\end{array}\right] = \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \end{gathered}$$

Equating the corresponding elements, we get 

$$\begin{gathered} 4{\beta }^2 + {\gamma }^2 = 1 .....\left(i\right) \\ 2{\beta }^2 - {\gamma }^2 = 0.......\left(ii\right) \end{gathered}$$

and ${\alpha }^2 + {\beta }^2 + {\gamma }^2 = 1 .....\left(iii\right)$

From Eqs. $\left(i\right)$and $\left(ii\right)$ we get, 

${\beta }^2 = \frac{1}{6}$and ${\gamma }^2 = \frac{1}{3}$

From Eq. $\left(iii\right)$

${\alpha }^2 = 1 - {\beta }^2-{\gamma }^2 = 1-\frac{1}{6} - \frac{1}{3} = \frac{1}{2}$

Hence, $2{\alpha }^2 + 6{\beta }^2 + 3{\gamma }^2 = 2 \times \frac{1}{2} + 6 \times \frac{1}{6} + 3 \times \frac{1}{3} = 3$

Aliter: 

The rows of matrix $A$ are unit orthogonal vectors 

$$\begin{gathered} \overrightarrow{R_1} . \overrightarrow{R_2} = 0 \Rightarrow 2{\beta }^2 - {\gamma }^2 = 0 \Rightarrow 2{\beta }^2 = {\gamma }^2 \\ \overrightarrow{R_2} . \overrightarrow{R_3} = 0 \Rightarrow {\alpha }^2 -{\beta }^2 -{\gamma }^2 =0 \Rightarrow {\beta }^2 + {\gamma }^2 = {\alpha }^2 \end{gathered}$$

and $\overrightarrow{R_2} .\overrightarrow{R_3 } =1 \Rightarrow {\alpha }^2 + {\beta }^2 + {\gamma }^2 = 1$

From Eqs $\left(i\right), \left(ii\right)$and $\left(iii\right)$, we get 

${\alpha }^2 = \frac{1}{2}, {\beta }^2 = \frac{1}{6}$and ${\gamma }^2 = \frac{1}{3}$

$\therefore 2{\alpha }^2 + 6{\beta }^2 + 3{\gamma }^2 = 3$