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

To decide whether $$\vec c$$ lies in the plane determined by $$\vec a$$ and $$\vec b$$, recall the following fact:
Three vectors $$\vec a, \vec b, \vec c$$ are coplanar ⇔ their scalar triple product vanishes, i.e. $$\vec a \cdot (\vec b \times \vec c)=0$$.

Write the components of the three given vectors:
$$\vec a = (1,\,1,\,1), \quad \vec b = (1,\,-1,\,2), \quad \vec c = (x,\,x-2,\,-1).$$

Form the scalar triple product as a determinant:
$$\vec a \cdot (\vec b \times \vec c)= \begin{vmatrix} 1 & 1 & 1\\ 1 & -1 & 2\\ x & x-2 & -1 \end{vmatrix}=0.$$ Evaluate the determinant (cofactor expansion along the first row):
$$\begin{aligned} &= 1\Big[(-1)(-1)-2(x-2)\Big] - 1\Big[1(-1)-2x\Big] + 1\Big[1(x-2)-(-1)x\Big] \\ &= 1\big[1-2x+4\big] - 1\big[-1-2x\big] + 1\big[(x-2)+x\big] \\ &= (5-2x) + (1+2x) + (2x-2) \\ &= 2x + 4. \end{aligned}$$

Set the result to zero for coplanarity:
$$2x + 4 = 0 \;\; \Longrightarrow \;\; x = -2.$$

Thus the required value of $$x$$ is $$-2$$.

Option D which is: $$-2$$