Let $\vec a=\hat i-\hat k,\,\vec b=x\hat i+\hat j+(1-x)\hat k$ and $\vec c=y\hat i+x\hat j+(1+x-y)\hat k$. Then $[\vec a\,\vec b\,\vec c]$ depends on:
Answer & explanation
Correct answer: option 1
$[\vec a\,\vec b\,\vec c]=\begin{bmatrix}1&0&-1\\x&1&1-x\\y&x&1+x-y\end{bmatrix}=(1+x-y-x+x^2)-(x^2-y)=1$