Given $\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k}, \overrightarrow{b}=\widehat{i}-\widehat{j}+2\widehat{k}$ and $\overrightarrow{c}=x\widehat{i}+\left(x-2\right)\widehat{j}-\widehat{k}.$

Since, the vector $\overrightarrow{c}$ lies in the plane of vectors $\overrightarrow{a}$ and $\overrightarrow{b},$ therefore $\overrightarrow{a}, \overrightarrow{b}$ and $\overrightarrow{c}$ are coplanar.

If three vectors $a_1\widehat{i}+a_2\widehat{j}+a_3\widehat{k}, b_1\widehat{i}+b_2\widehat{j}+b_3\widehat{k}$ and $c_1\widehat{i}+c_2\widehat{j}+c{}_3\widehat{k}$ are coplanar, then

$\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|=0.$

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

$\Rightarrow 1\times \left(1-2x+4\right)-1\times \left(-1-2x\right)+1\times \left(x-2+x\right)=0$

$\Rightarrow 5-2x+1+2x+2x-2=0$

$\Rightarrow x=-2$