Step 1: Simplify with column operations:
Apply $C_3 \to C_3 + C_1$. The new third column has entries $-1 + 1 = 0$, $(1 - x) + x = 1$ and $(1 + x - y) + y = 1 + x$.
Step 2: Expand:
The determinant is $\begin{vmatrix} 1 & 0 & 0 \\ x & 1 & 1 \\ y & x & 1 + x\end{vmatrix} = 1\cdot[(1)(1 + x) - (1)(x)] = 1$.
A constant, so neither $x$ nor $y$ matters.
Final Answer:
The value is $1$, independent of $x$ and $y$, option (B).
\[ \boxed{\text{neither } x \text{ nor } y} \]