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 $[x~y~z]$ be the third row. Due to the orthogonality, we have $$\frac{x}{3}+\frac{2y}{3}+\frac{2z}{3}=0,$$ and $$\frac{y}{\sqrt{2}}-\frac{z}{\sqrt{2}}=0.$$ Consequently, $y=z$, and $x=-4y$. Using $x^2+y^2+z^2=1$, the values of $x$, $y$, and $z$ can easily be found. The last condition ensures that the vector has unit norm. 
If there is no constraint on the norm of the third vector, there will be infinite solutions. One such is $[-4~1~1]$. But, if the norm of the third vector is 1 (as is the case with the other two vectors), there are only two such solutions, which are given below.
$$\begin{bmatrix}\frac{-4}{\sqrt{18}} & \frac{1}{\sqrt{18}} & \frac{1}{\sqrt{18}}\end{bmatrix},$$ and $$\begin{bmatrix}\frac{4}{\sqrt{18}} & \frac{-1}{\sqrt{18}} & \frac{-1}{\sqrt{18}}\end{bmatrix}.$$
