Let $$\vec a = \hat i + \hat j + \hat k,\,\vec b = \hat i - \hat j + 2\hat k$$       and $$\vec c = x\hat i + \left( {x - 2} \right)\hat j - \hat k.$$     If the vectors $${\vec c}$$ lies in the plane of $${\vec a}$$ and $${\vec b},$$  then $$x$$ equals :

Solution :
Given $$\vec a = \hat i + \hat j + \hat k,\,\vec b = \hat i - \hat j + 2\hat k$$       and $$\vec c = x\hat i + \left( {x - 2} \right)\hat j - \hat k$$
If $${\vec c}$$ lies in the plane of $${\vec a}$$ and $${\vec b},$$  then $$\left[ {\vec a\,\vec b\,\vec c} \right] = 0$$
\[\begin{array}{l} {\rm{i}}{\rm{.e}}{\rm{.,}}\,\left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\\ x\,\,\,\,\left( {x - 2} \right)\,\,\,\, - 1 \end{array} \right| = 0\\ \Rightarrow 1\left[ {1 - 2\left( {x - 2} \right) - 1\left[ { - 1 - 2x} \right]} \right] + 1\left[ {x - 2 + x} \right] = 0\\ \Rightarrow 1 - 2x + 4 + 1 + 2x + 2x - 2 = 0\\ \Rightarrow 2x = - 4\\ \Rightarrow x = - 2 \end{array}\]