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}
 A: One of the possible solutions:
$$\frac{\begin{pmatrix} \frac{1}{3} \\\frac{2}{3} \\ \frac{2}{3}\end{pmatrix} \times \begin{pmatrix} 0 \\\frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}}\end{pmatrix}}{\left\Vert\begin{pmatrix} \frac{1}{3} \\\frac{2}{3} \\ \frac{2}{3}\end{pmatrix} \times \begin{pmatrix} 0 \\\frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}}\end{pmatrix}\right\Vert}$$
Why this solution works:
An othogonal matrix has an orthonormal basis in its rows, so by taking the cross product of the two given vectors, you assure that your vector is perpendicular to both given vectors. We then devide the vector by its own norm, because then it has length $1$. Therefore, together with the two given vectors this vector will form an orthonormal basis. Note that the inverse (for addition) of this vector will also be a correct solution.
There are also other possibilities to solve this question. You can solve it by inspection, but this requires some insight. Another way would be to use the Gramm-Schmidt algorithm.
A: 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 $x_2 = \frac{4}{\sqrt{18}}, y_2 = -\frac{1}{\sqrt{18}}, z_2 = -\frac{1}{\sqrt{18}}$.
A: 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 $$\begin{bmatrix}\frac{4}{\sqrt{18}} & \frac{-1}{\sqrt{18}} & \frac{-1}{\sqrt{18}}\end{bmatrix}.$$
A: 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 $a.$
$(\frac 13,\frac 23,\frac23)\cdot(a,3,3) = 0\\
\frac {a}{3} + 4 = 0\\
(-12,3,3)$
