The value of x so that the four points  $A=(0,2,0)$ $\mathrm{B}=(1,\mathrm{x},0),\mathrm{C}=(1,2,0)\text{ and }\mathrm{D}=(1,2,1)$ are coplanar, is:

Let $\overrightarrow{A}=2\widehat{j};\overrightarrow{B}=\widehat{i}+x\widehat{j},\overrightarrow{\mathrm{C}}=\widehat{i}+2\widehat{j}$ and $\overrightarrow{\mathrm{D}}=\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+\widehat{\mathrm{k}}$

So $$\begin{gathered} \overrightarrow{\mathrm{AB}}=\widehat{\mathrm{i}}+(\mathrm{x}-2)\widehat{\mathrm{j}} \\ \begin{array}{r}\overrightarrow{\mathrm{AC}}=+\widehat{\mathrm{i}}\\ \overrightarrow{\mathrm{AD}}=\widehat{\mathrm{i}}+\widehat{\mathrm{k}}\end{array} \end{gathered}$$

If $\overrightarrow{\mathrm{A}},\overrightarrow{\mathrm{B}},\overrightarrow{\mathrm{C}},\overrightarrow{\mathrm{D}}$ are coplanar vectors then $\overrightarrow{\mathrm{AB}},\overrightarrow{\mathrm{AC}}\text{ and }\overrightarrow{\mathrm{AD}}$ are also coplanar vectors. 

$\left|\begin{array}{ccc}1& x-2& 0\\ 1& 0& 0\\ 1& 0& 1\end{array}\right|=0$

$\begin{array}{l}\Rightarrow -(x-2)(1-0)=0\\ \Rightarrow x-2=0\end{array}$

Thus $x=2$.