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

The scalar triple product of three vectors is defined as
$$[\vec{a},\vec{b},\vec{c}] \;=\; \vec{a}\cdot(\vec{b}\times\vec{c})$$
and is equal to the determinant whose rows (or columns) are the components of the three vectors.

The given vectors are
$$\vec{a}=1\,\hat{i}+0\,\hat{j}-1\,\hat{k}\;=\;(1,\,0,\,-1)$$
$$\vec{b}=x\,\hat{i}+1\,\hat{j}+(1-x)\,\hat{k}\;=\;(x,\,1,\,1-x)$$
$$\vec{c}=y\,\hat{i}+x\,\hat{j}+(1+x-y)\,\hat{k}\;=\;(y,\,x,\,1+x-y)\,.$$

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

Expand the determinant along the first row:

$$ \begin{aligned} [\vec{a},\vec{b},\vec{c}] &= 1\;\begin{vmatrix} 1 & 1-x\\ x & 1+x-y \end{vmatrix} \;-\;0\;\begin{vmatrix} x & 1-x\\ y & 1+x-y \end{vmatrix} \;+\;(-1)\;\begin{vmatrix} x & 1\\ y & x \end{vmatrix}\\[4pt] &= 1\;\bigl[(1)(1+x-y)-x(1-x)\bigr] \;-\;\bigl[x(x)-y(1)\bigr]\\[4pt] &= 1\;\bigl[1+x-y -x + x^{2}\bigr] \;-\;\bigl[x^{2}-y\bigr]\\[4pt] &= 1\;\bigl[1 + x^{2} - y\bigr] \;-\;x^{2} + y\\[4pt] &= 1 + x^{2} - y - x^{2} + y\\[4pt] &= 1. \end{aligned} $$

The final value of the scalar triple product is the constant $$1$$, which does not involve either $$x$$ or $$y$$. Therefore $$[\vec{a},\vec{b},\vec{c}]$$ is independent of both variables.

Option D which is: neither $$x$$ nor $$y$$