You may use the augmented matrix method to find the inverse matrix.
Here we give an alternative way to find the inverse matrix by noting that $A$ is an orthogonal matrix.

Recall that a matrix $B$ is orthogonal if $B^{\trans}B=B^{\trans}B=I$.
Thus, once we know $B$ is an orthogonal matrix, then the inverse matrix $B^{-1}$ is just the transpose matrix $B^{\trans}$.

Also, recall that a matrix $B$ is orthogonal if and only if the column vectors of $B$ form an orthonormal set.

Solution.

We first show that $A$ is an orthogonal matrix.
To do this, it suffices to that the column vectors form an orthonormal set.

Let
\[ \mathbf{v}_1= \begin{bmatrix}
\frac{2}{7} \\[6 pt]
\frac{6}{7} \\[6pt]
-\frac{3}{7}
\end{bmatrix},
\mathbf{v}_2=\begin{bmatrix}
\frac{3}{7} \\[6 pt]
\frac{2}{7} \\[6pt]
\frac{6}{7} \end{bmatrix},
\mathbf{v}_3=\begin{bmatrix}
\frac{6}{7} \\[6 pt]
-\frac{3}{7} \\[6pt]
-\frac{2}{7}
\end{bmatrix}\]
be the column vectors of $A$.

Then the length of the vector $\mathbf{v}_1$ is
\[||\mathbf{v}_1||=\sqrt{(2/7)^2+(6/7)^2+(-3/7)^2}=1.\]
Similarly, we have $||\mathbf{v}_2||=||\mathbf{v}_3||=1$.
Thus, column vectors are unit vectors.

The dot (inner) product of the vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ is
\[\mathbf{v}_1\cdot \mathbf{v}_2=\frac{2}{7}\cdot \frac{3}{7}+\frac{6}{7}\cdot \frac{2}{7}+\left( -\frac{3}{7}\right) \cdot \frac{6}{7}=0.\]
Similarly, we have
\[\mathbf{v}_1\cdot \mathbf{v}_3=0, \quad \mathbf{v}_2\cdot \mathbf{v}_3=0.\]

Therefore, the column vectors are orthogonal.
Hence the column vectors of $A$ are orthonormal, and this implies that $A$ is an orthogonal matrix. Namely, $A^{\trans}=A^{-1}$.
Thus the inverse matrix of $A$ is
\[A^{-1}=\begin{bmatrix}
\frac{2}{7} & \frac{6}{7} & -\frac{3}{7} \\[6 pt]
\frac{3}{7} &\frac{2}{7} &\frac{6}{7} \\[6pt]
\frac{6}{7} & -\frac{3}{7} & -\frac{2}{7}
\end{bmatrix}.\]

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