Step-by-step Solution:

To solve the problem, we need to analyze the determinant of the matrix formed by the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). Specifically, we determine whether this determinant depends on the variables \(x\) and \(y\).\[\] Step 1: Write the vectors in matrix form The given vectors are: \[ \vec{a} = \hat{i} - \hat{k}, \quad \vec{b} = x\hat{i} + \hat{j} + (1-x)\hat{k}, \quad \vec{c} = y\hat{i} + x\hat{j} + (1+x-y)\hat{k}. \] We form the matrix: \[ M = \begin{bmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1-x & 1+x-y \end{bmatrix}. \] \[\] Step 2: Calculate the determinant The determinant of \(M\) is: \[ \text{Det}(M) = \begin{vmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1-x & 1+x-y \end{vmatrix}. \] Using the cofactor expansion along the first row: \[ \text{Det}(M) = 1 \cdot \begin{vmatrix} 1 & x \\ 1-x & 1+x-y \end{vmatrix} - x \cdot \begin{vmatrix} 0 & x \\ -1 & 1+x-y \end{vmatrix} + y \cdot \begin{vmatrix} 0 & 1 \\ -1 & 1-x \end{vmatrix}. \] \[\] Step 3: Compute each minor determinant 1. First minor: \[ \begin{vmatrix} 1 & x \\ 1-x & 1+x-y \end{vmatrix} = (1)(1+x-y) - (x)(1-x) = 1 + x - y - x + x^2 = 1 + x^2 - y. \] 2. Second minor: \[ \begin{vmatrix} 0 & x \\ -1 & 1+x-y \end{vmatrix} = (0)(1+x-y) - (x)(-1) = x. \] 3. Third minor: \[ \begin{vmatrix} 0 & 1 \\ -1 & 1-x \end{vmatrix} = (0)(1-x) - (1)(-1) = 1. \] \[\] Step 4: Substitute back into the determinant \[ \text{Det}(M) = 1 \cdot (1 + x^2 - y) - x \cdot x + y \cdot 1. \] Simplify: \[ \text{Det}(M) = 1 + x^2 - y - x^2 + y. \] \[ \text{Det}(M) = 1. \] \[\] Step 5: Conclusion The determinant of the matrix \(\text{Det}(M)\) simplifies to a constant value, \(1\), and does not depend on \(x\) or \(y\). \[\] Final Answer: Option A: Neither \(x\) nor \(y\).