Since the given matrix is orthogonal, we can write ,

$A.A^{'}= I \\ \therefore \frac1{81} \begin{bmatrix} -8 & 4 & a \\ 1 & 4 & b\\ 4 & 7 & c \end{bmatrix} . \begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ a & b & c \end{bmatrix} = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \\ \\ \therefore {64 +16+ a^2 \over 81} = 1\\ \therefore a^2 = 81-80 \\ \therefore a^2 = 1 \\ \therefore a = \pm 1 \\ \therefore {1+16+b^2 \over 81}=1 \\ \therefore b^2 = 81-17 \\ \therefore b^2 = 64 \\ \therefore b = \pm 8 \\ \therefore {16+49+c^2 \over 81} = 1 \\ \therefore c^2 = 81-65 \\ \therefore c^2 = 16 \\ \therefore c = \pm 4 \\ \therefore A = \frac19 \begin{bmatrix} -8 & 4 & \pm1 \\ 1 & 4 & \pm 8 \\ 4 & 7 & \pm4 \end{bmatrix}\\ \\ $

According to the property of orthogonal matrices, if A is an orthogonal matrix then,

$A^{-1} \; exists \; and \; is \; equal \; to \; A^{'} \\ \therefore A^{-1} = \frac19 \begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ \pm1 & \pm 8 & \pm 4 \end{bmatrix}$