This problem tests your understanding of determinant properties, especially row and column operations, and how to effectively use given algebraic conditions to simplify expressions. The key is to strategically apply operations to create zeros and common factors, which greatly simplifies the determinant calculation.

Key Concepts and Formulas

1. Properties of Determinants:
   • Row/Column Operations: The value of a determinant remains unchanged if we apply the operation $R_i \to R_i + k R_j$ (or $C_i \to C_i + k C_j$), where $R_i$ (or $C_i$) denotes the $i$-th row (or column). This is invaluable for simplifying determinants by introducing zeros.
   • Factoring Common Elements: If all elements of a row or column have a common factor, it can be taken out of the determinant.
   • Expansion of a Determinant: A $3 \times 3$ determinant can be expanded along any row or column. For an element $a_{ij}$, its cofactor is $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor obtained by deleting the $i$-th row and $j$-th column. The determinant is the sum of the products of elements of any row/column with their corresponding cofactors.

Step-by-Step Solution

Let the given determinant be $D$. D = \left| {\matrix{ x & {a + y} & {x + a} \cr y & {b + y} & {y + b} \cr z & {c + y} & {z + c} \cr } } \right|

Step 1: Simplify the Third Column ($C_3 \to C_3 - C_1$)

• Why this step? We observe that the elements in the third column ($x+a$, $y+b$, $z+c$) contain the corresponding elements of the first column ($x$, $y$, $z$). Subtracting $C_1$ from $C_3$ will eliminate $x, y, z$ from the third column, leaving behind simpler terms ($a, b, c$). This makes the determinant easier to manage.

Applying the operation $C_3 \to C_3 - C_1$: D = \left| {\matrix{ x & {a + y} & {(x+a) - x} \cr y & {b + y} & {(y+b) - y} \cr z & {c + y} & {(z+c) - z} \cr } } \right| D = \left| {\matrix{ x & {a + y} & a \cr y & {b + y} & b \cr z & {c + y} & c \cr } } \right|

Step 2: Simplify the Second Column ($C_2 \to C_2 - C_3$)

• Why this step? Now that the third column contains $a, b, c$, we notice that the second column elements are $a+y, b+y, c+y$. By subtracting $C_3$ from $C_2$, we can eliminate $a, b, c$ from the second column, leaving a common term '$y$' in each element. This creates a common factor, which is highly beneficial for further simplification.

Applying the operation $C_2 \to C_2 - C_3$: D = \left| {\matrix{ x & {(a+y) - a} & a \cr y & {(b+y) - b} & b \cr z & {(c+y) - c} & c \cr } } \right| D = \left| {\matrix{ x & y & a \cr y & y & b \cr z & y & c \cr } } \right|

Step 3: Create Zeros in the Second Column ($R_2 \to R_2 - R_1$, $R_3 \to R_3 - R_1$)

• Why this step? To efficiently evaluate a determinant, it's best to expand it along a row or column that contains the maximum number of zeros. Here, the second column has a common element '$y$'. By subtracting $R_1$ from $R_2$ and $R_3$, we can make the elements $a_{22}$ and $a_{32}$ zero, leaving only one non-zero element in $C_2$.

First, apply $R_2 \to R_2 - R_1$: D = \left| {\matrix{ x & y & a \cr {y - x} & {y - y} & {b - a} \cr z & y & c \cr } } \right| = \left| {\matrix{ x & y & a \cr {y - x} & 0 & {b - a} \cr z & y & c \cr } } \right| Next, apply $R_3 \to R_3 - R_1$: D = \left| {\matrix{ x & y & a \cr {y - x} & 0 & {b - a} \cr {z - x} & {y - y} & {c - a} \cr } } \right| D = \left| {\matrix{ x & y & a \cr {y - x} & 0 & {b - a} \cr {z - x} & 0 & {c - a} \cr } } \right|

Step 4: Expand the Determinant along the Second Column ($C_2$)

• Why this step? We have successfully created two zeros in the second column. Expanding along this column will involve only one term, simplifying the calculation significantly. Remember the sign convention for expansion: for an element $a_{ij}$, its cofactor has a sign of $(-1)^{i+j}$. For the element $y$ at position $(1,2)$, the sign factor is $(-1)^{1+2} = -1$.

Expanding $D$ along $C_2$: D = y \cdot (-1)^{1+2} \left| {\matrix{ {y - x} & {b - a} \cr {z - x} & {c - a} \cr } } \right| + 0 \cdot C_{22} + 0 \cdot C_{32} $D = -y \left[ (y - x)(c - a) - (b - a)(z - x) \right]$

Step 5: Utilize the Given Condition and Substitute

• Why this step? The problem provides a crucial relationship between $a, b, c, x, y, z$: $a + x = b + y = c + z + 1$. We must use this condition to simplify the expression and match one of the options. Let's denote the common value by $k$: $a + x = k \implies x = k - a$ $b + y = k \implies y = k - b$ $c + z + 1 = k \implies z = k - c - 1$

Now, substitute these expressions into the terms $(y-x)$ and $(z-x)$:

• $y - x = (k - b) - (k - a) = k - b - k + a = a - b$
• $z - x = (k - c - 1) - (k - a) = k - c - 1 - k + a = a - c - 1$

Substitute these simplified terms back into the determinant expression: $D = -y \left[ (a - b)(c - a) - (b - a)(a - c - 1) \right]$

Step 6: Final Simplification

• Why this step? We need to simplify the algebraic expression to match one of the given options. Notice that $(b-a)$ is the negative of $(a-b)$. We can use this to factor out a common term.

Replace $(b-a)$ with $-(a-b)$: $D = -y \left[ (a - b)(c - a) - (-(a - b))(a - c - 1) \right]$ $D = -y \left[ (a - b)(c - a) + (a - b)(a - c - 1) \right]$ Now, factor out $(a-b)$ from the bracketed expression: $D = -y (a - b) \left[ (c - a) + (a - c - 1) \right]$ $D = -y (a - b) \left[ c - a + a - c - 1 \right]$ $D = -y (a - b) \left[ -1 \right]$ $D = y (a - b)$

This result matches option (B).

Common Mistakes & Tips:

• Sign Errors in Expansion: Be very careful with the signs when expanding a determinant, especially the $(-1)^{i+j}$ factor.
• Algebraic Simplification: Double-check your algebraic manipulations, especially when dealing with negative signs and factoring.
• Using Given Conditions: Always remember to use all given conditions in the problem. They are usually crucial for simplifying the expression to one of the options.
• Strategic Operations: Choose row/column operations that genuinely simplify the determinant (e.g., by creating zeros or common factors). Don't just apply operations randomly.

Final Answer: The value of the determinant is $y(a-b)$.

The final answer is $\boxed{\text{y(a – b)}}$.