<s>[INST] 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}
 [/INST] One approach that would be to find the cross product.
$(\frac 13, \frac 23, \frac 23)\times (0, \frac 1{\sqrt 2},-\frac 1 {\sqrt2}) = (-\frac 4{3\sqrt 2},\frac {1}{3\sqrt2},\frac {1}{3\sqrt2})$
Another would be to say by inspection any vector that is orthogonal to the second vector must have the from $(a,b,b)$
Choose an arbitrary value for $b$ (3 is convenient) and solve for $a.$
$(\frac 13,\frac 23,\frac23)\cdot(a,3,3) = 0\\
\frac {a}{3} + 4 = 0\\
(-12,3,3)$
</s>