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

neither x nor y

We have,

$[\vec a\,\vec b\,\vec c]=\begin{vmatrix}1&0&-1\\x&1&1-x\\y&x&1+x-y\end{vmatrix}$

$⇒[\vec a\,\vec b\,\vec c]=\begin{vmatrix}1&0&0\\x&1&1\\y&x&1+x\end{vmatrix}$  Applying $C_3→ C_3 +C_1$

$⇒[\vec a\,\vec b\,\vec c]=1$, which is independent of both x and y