The vectors $\vec a =x\hat i +(x+1)\hat j+(x+2)\hat k, \vec b=(x+3)\hat i+(x+4)\hat j+(x+5)\hat k$ and, $\vec c=(x+6)\hat i+(x+7)\hat j+(x+8)\hat k$ are coplanar for
Answer & explanation
Correct answer: option 1
Vectors $\vec a,\vec b,\vec c$ will be coplanar, iff $[\vec a\,\vec b\,\vec c]=0$
We have,
$\begin{bmatrix}\vec a&\vec b&\vec c\end{bmatrix}=\begin{vmatrix}x&x+1&x+2\\x+3&x+4&x+5\\x+6&x+7&x+8\end{vmatrix}$
$⇒\begin{bmatrix}\vec a&\vec b&\vec c\end{bmatrix}=\begin{vmatrix}x&x+1&x+2\\3&3&3\\6&6&6\end{vmatrix}$ [Applying $R_2 → R_2-R_1, R_3 → R_3-R_1]$
$⇒\begin{bmatrix}\vec a&\vec b&\vec c\end{bmatrix}=0$ for all x [$∵R_1$ and $R_2$ are proportional]
Hence, vectors are coplanar for all values of x.