Find 3×3 orthogonal matrix satisfying the following conditions. Let $A$ be 3×3 orthogonal real matrix.
For $\mathbf{u}$ $= \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \mathbf{v}$ $= \begin{pmatrix} 0 \\ 0 \\ \sqrt3 \end{pmatrix}, \mathbf{w}$ $= \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$,
$Au=v, Aw=-w$ and $detA=-1$
Find $A$.
I noted that $u, v, w$ are linearly independent. And I write
$A\begin{pmatrix}
  1 & 0 & 1 \\
  1 & 0 & -1 \\
  1 & \sqrt3 & 0
 \end{pmatrix}=\begin{pmatrix}
  0 & a & -1 \\
  0 & b & 1 \\
  \sqrt3 & c & 0
 \end{pmatrix}.$
Then $A=\begin{pmatrix}
  0 & a & -1 \\
  0 & b & 1 \\
  \sqrt3 & c & 0
 \end{pmatrix}\begin{pmatrix}
  \frac{1}{2} & \frac{1}{2} & 0 \\
  -\frac{1}{2\sqrt3} & -\frac{1}{2\sqrt3} & \frac{1}{\sqrt3} \\
  \frac{1}{2} & -\frac{1}{2} & 0
 \end{pmatrix}$
Now, how can I find $a, b, c$??
Am I doing this right?
If there is a better solution, please help me. 

Hint
As $A$ is an orthogonal matrix it preserves the inner product. Therefore $\Vert Av\Vert = \sqrt 3$. Also $\langle Av, Aw\rangle =\langle v, w\rangle=0$.
Finally use the equation $\det A =-1$.
This provides 3 equations to find $a,b,c$.
