Finding third row of orthogonal matrix? 
Find a $3\times3$ orthogonal matrix whose first two rows are $\Big[\frac{1}{3},\frac{2}{3},\frac{2}{3}\Big]$ and $\left[0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right]$.

I tried two approaches.
One, finding vector cross product of given two rows.
Second, assuming third row as $[x,y,z]$ and applying the property of orthogonal matrix.
In each approach I got a different solution, which is not correct.
Even I suspect there is an error in given problem.
Please help me in finding the third row.
Adding cross product calculation below.
\begin{vmatrix}
\bar{\imath}&\bar{\jmath}&\bar{k}\\\frac{1}{3}&\frac{2}{3}&\frac{2}{3}\\0&\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}
\end{vmatrix}
\begin{align}
&=\bar{\imath}(0)-\bar{\jmath}\left(-\frac{1}{3\sqrt{2}}\right)+\bar{k}\left(\frac{1}{3\sqrt{2}}\right) \\[1ex]
&=\frac{1}{\sqrt{\frac{1}{18}+\frac{1}{18}}}\left(0,\frac{1}{3\sqrt{2}},\frac{1}{3\sqrt{2}}\right) \\[1ex]
&=\left(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)
\end{align}

Let this matrix be denoted as:
\begin{equation}
\mathbf{Q}=\left[\begin{array}{ccc}
\frac{1}{3} & \frac{2}{3} & \frac{2}{3}\\
0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\
x & y & z
\end{array}\right]
\end{equation}
From the definition of the orthogonal matrix $\mathbf{Q}\,\mathbf{Q}^T = \mathbf{I}$:
\begin{equation}
\left[\begin{array}{ccc}
\frac{1}{3} & \frac{2}{3} & \frac{2}{3}\\
0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\
x & y & z
\end{array}\right]
\left[\begin{array}{ccc}
\frac{1}{3} & 0 & x\\
\frac{2}{3} & \frac{1}{\sqrt{2}} & y\\
\frac{2}{3} & -\frac{1}{\sqrt{2}} & z
\end{array}\right]
=
\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}\right]
\end{equation}
we get three equations:
\begin{align}
&\frac{1}{3}x + \frac{2}{3}y + \frac{2}{3}z = 0\\
&\frac{1}{\sqrt{2}}y - \frac{1}{\sqrt{2}}z = 0\\
&x^2 + y^2 + z^2 = 1
\end{align}
By solving this you get two solutions $x_1 = -\frac{4}{\sqrt{18}}, y_1 = \frac{1}{\sqrt{18}}, z_1 = \frac{1}{\sqrt{18}}$ or $x_2 = \frac{4}{\sqrt{18}}, y_2 = -\frac{1}{\sqrt{18}}, z_2 = -\frac{1}{\sqrt{18}}$.
