If the vectors $\vec{a}=\hat{i}+\hat{j}+\hat{k},\vec{b}=\hat{i}-\hat{j}+2\hat{k},$ and $\vec{c}=x\hat{i}+(x-2)\hat{j}+\hat{k}$ are coplanar, then the value of $x$ is:

Solution

1. The condition for coplanarity is the determinant of the components being zero: $\left|\begin{array}{ccc}1& 1& 1\\ 1& -1& 2\\ x& x-2& 1\end{array}\right|=0$.
2. Expand the determinant: $1(-1 - 2(x-2)) - 1(1 - 2x) + 1(x-2 - (-x)) = 0$.
3. Simplify the expression: $(-1-2x+4)-(1-2x)+(2x-2)=0$.
4. Combine terms: $3 - 2x - 1 + 2x + 2x - 2 = 0$, which simplifies to $2x = 4$, so $x=2$.
5. Hence the answer is (B).