$$ A=\dfrac 19 \begin{vmatrix} -8 & 4& a\ 1&4&b\ 4&7&c\ \end{vmatrix} $$

Taking transpose

$A^{-1} =\dfrac 19 \begin{vmatrix} -8&1&4\\ 4&4&7\\ a&b&c \\ \end{vmatrix} $

Since A is orthogonal

We know $AA^1=I$

$\therefore \dfrac 19 \begin{vmatrix} -8 & 4& a\\ 1&4&b\\ 4&7&c\\ \end{vmatrix}\times \dfrac 19 \begin{vmatrix} -8&1&4\\ 4&4&7\\ a&b&c \\ \end{vmatrix} =\begin{vmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \\ \end{vmatrix} \\ \therefore \dfrac 1{81} \begin{vmatrix} 64+16+a^2&-8+16+ab&-32+28+ac\\ -8+16+ab&1+16+b^2&4+28+bc\\ -32+28+ac&4+28+bc&16+49+c^2 \\ \end{vmatrix}=\begin{vmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \\ \end{vmatrix} \\ \therefore \begin{vmatrix} 80+a^2&8+ab&-4+ac\\ 8+ab&17+b^2&32+bc\\ -4+ac&32+bc&65+c^2 \\ \end{vmatrix}=\begin{vmatrix} 81&0&0\\ 0&81&0\\ 0&0&81 \\ \end{vmatrix} $

Comparing both sides we get

$80+a^2=81 \space\space\space\space ∴a^2=1 \space\space\space a=±1 \\ 8+ab=0 \space\space\space ∴ ab=-8 \\ ∴b=\dfrac {-8}9 \\ ∴b=∓8 \\ -4+ac=0 \space\space\space\space ∴ac=4 \space\space\space\space ∴c=\dfrac 4a \\ ∴c=±4 $

∴ Value of $a, b, c$ are $±1 ,∓8,±4$ respectively.