Determining the last column so that the resulting matrix is an orthogonal matrix 
Determine the last column so that the resulting matrix is an orthogonal
  matrix $$\begin{bmatrix} \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{6}} & ? \\ \dfrac{1}{\sqrt{2}} & -\dfrac{1}{\sqrt{6}} & ? \\ 0 & \dfrac{2}{\sqrt{6}} & ? \end{bmatrix}$$

Can anyone please provide hints to solve this?

A matrix is orthogonal if all the column vectors are unit vectors and any two columns have dot product zero. Now write the unknown last column as $(x\ y\ z)^T$.  You will get two equations when when you insist it be orthogonal to the known first and second columns.
So solve a system of 2 equations an $x,y,z$. There will be infinitely many solutions. Now the condition on length will bring down the solutions to 2 (a vector and its negative).
