Null space of $$$\left[\begin{array}{ccc}\frac{\sqrt{6}}{6} & \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{6}\\- \frac{\sqrt{3}}{3} & 0 & \frac{\sqrt{6}}{3}\end{array}\right]$$$
Your Input
Find the null space of $$$\left[\begin{array}{ccc}\frac{\sqrt{6}}{6} & \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{6}\\- \frac{\sqrt{3}}{3} & 0 & \frac{\sqrt{6}}{3}\end{array}\right]$$$.
Solution
The reduced row echelon form of the matrix is $$$\left[\begin{array}{ccc}1 & 0 & - \sqrt{2}\\0 & 1 & 1\end{array}\right]$$$ (for steps, see rref calculator).
To find the null space, solve the matrix equation $$$\left[\begin{array}{ccc}1 & 0 & - \sqrt{2}\\0 & 1 & 1\end{array}\right]\left[\begin{array}{c}x_{1}\\x_{2}\\x_{3}\end{array}\right] = \left[\begin{array}{c}0\\0\end{array}\right].$$$
If we take $$$x_{3} = t$$$, then $$$x_{1} = \sqrt{2} t$$$, $$$x_{2} = - t$$$.
Thus, $$$\mathbf{\vec{x}} = \left[\begin{array}{c}\sqrt{2} t\\- t\\t\end{array}\right] = \left[\begin{array}{c}\sqrt{2}\\-1\\1\end{array}\right] t.$$$
This is the null space.
The nullity of a matrix is the dimension of the basis for the null space.
Thus, the nullity of the matrix is $$$1$$$.
Answer
The basis for the null space is $$$\left\{\left[\begin{array}{c}\sqrt{2}\\-1\\1\end{array}\right]\right\}\approx \left\{\left[\begin{array}{c}1.414213562373095\\-1\\1\end{array}\right]\right\}.$$$A
The nullity of the matrix is $$$1$$$A.