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

-2

It is given that $\vec a,\vec b,\vec c$ are coplanar.

$∴[\vec a\,\vec b\,\vec c]=0$

$⇒\begin{vmatrix}1&1&1\\1&-1&2\\x&x-2&-1\end{vmatrix}=0$

$⇒(1-2x+4)-(-1-2x) + (x-2+x) = 0$

$⇒2x+4=0⇒ x=-2$