1. Key Concepts and Formulas

• Row/Column Operations: The value of a determinant remains unchanged if we apply operations of the form $R_i \to R_i + kR_j$ (adding a scalar multiple of one row to another row) or $C_i \to C_i + kC_j$ (adding a scalar multiple of one column to another column). These operations are fundamental for simplifying determinants.
• Factoring Common Terms: If all elements of a particular row or column share a common factor, this factor can be taken out of the determinant. For example, if $k$ is common in $R_i$, then $\det(A) = k \det(A')$, where $A'$ is the matrix with $R_i$ divided by $k$.
• Expansion of Determinants: A $3 \times 3$ determinant can be expanded along any row or column. Expanding along a row/column that contains zeros significantly reduces the number of terms to calculate, simplifying the process.

2. Step-by-Step Solution

We are given the determinant: \Delta = \left| {\matrix{ {a - b - c} & {2a} & {2a} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right| And it is given that $\Delta = (a + b + c) (x + a + b + c)^2$, with $x \ne 0$. Our objective is to find the value of $x$.

Step 1: Create a Common Factor in the First Row

• Action: Apply the row operation $R_1 \to R_1 + R_2 + R_3$.
• Reasoning: This operation is chosen strategically to simplify the elements in the first row. By adding the corresponding elements from the second and third rows to the first row, we aim to reveal a common factor.
  • First element: $(a - b - c) + 2b + 2c = a + b + c$
  • Second element: $2a + (b - c - a) + 2c = a + b + c$
  • Third element: $2a + 2b + (c - a - b) = a + b + c$

Applying the operation, the determinant becomes: \Delta = \left| {\matrix{ {a + b + c} & {a + b + c} & {a + b + c} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right|

Step 2: Factor out the Common Term

• Action: Take $(a + b + c)$ common from the first row ($R_1$).
• Reasoning: Factoring out the common term simplifies the determinant matrix, making subsequent calculations easier and bringing us closer to the target form involving $(a+b+c)$.

After factoring, the determinant is: \Delta = (a + b + c) \left| {\matrix{ {1} & {1} & {1} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right|

Step 3: Create Zeros in the First Row

• Action: Apply the column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$.
• Reasoning: Having a '1' in the first row provides an excellent opportunity to create zeros in that row. Creating zeros simplifies the determinant's expansion considerably, as we will only need to calculate a smaller $2 \times 2$ determinant.

Let's perform the operations:

• For $C_2 \to C_2 - C_1$:
  • $1 - 1 = 0$
  • $(b - c - a) - 2b = -b - c - a = -(a + b + c)$
  • $2c - 2c = 0$
• For $C_3 \to C_3 - C_1$:
  • $1 - 1 = 0$
  • $2b - 2b = 0$
  • $(c - a - b) - 2c = -c - a - b = -(a + b + c)$

The determinant now looks like this: \Delta = (a + b + c) \left| {\matrix{ {1} & {0} & {0} \cr {2b} & {-(a + b + c)} & {0} \cr {2c} & {0} & {-(a + b + c)} \cr } } \right|

Step 4: Expand the Determinant

• Action: Expand the determinant along the first row ($R_1$).
• Reasoning: Since the first row now contains two zeros, expanding along this row will be very straightforward, involving only one non-zero term.

Expanding along $R_1$: \Delta = (a + b + c) \left[ 1 \cdot \left| {\matrix{ {-(a + b + c)} & {0} \cr {0} & {-(a + b + c)} \cr } } \right| - 0 \cdot (\text{minor}) + 0 \cdot (\text{minor}) \right] $\Delta = (a + b + c) \left[ (-(a + b + c)) \cdot (-(a + b + c)) - (0 \cdot 0) \right]$ $\Delta = (a + b + c) \left[ (a + b + c)^2 - 0 \right]$ $\Delta = (a + b + c) (a + b + c)^2$ $\Delta = (a + b + c)^3$

Step 5: Equate and Solve for x

We are given that $\Delta = (a + b + c) (x + a + b + c)^2$. From our calculation, we found $\Delta = (a + b + c)^3$.

Equating the two expressions: $(a + b + c)^3 = (a + b + c) (x + a + b + c)^2$

Assuming $a + b + c \ne 0$, we can divide both sides by $(a + b + c)$: $(a + b + c)^2 = (x + a + b + c)^2$

Taking the square root of both sides, we get two possibilities:

1. $a + b + c = x + a + b + c$ This implies $x = 0$. However, the problem statement explicitly states that $x \ne 0$. Therefore, this solution is not valid.

2. $a + b + c = -(x + a + b + c)$ $a + b + c = -x - a - b - c$ Rearranging the terms to solve for $x$: $x = -a - b - c - (a + b + c)$ $x = -2a - 2b - 2c$ $x = -2(a + b + c)$

This value of $x$ is consistent with the condition $x \ne 0$ (unless $a+b+c=0$, in which case the original equation would be $0=0$, but for a general solution, we consider $a+b+c \ne 0$).

3. Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when performing row/column operations (especially subtractions) and when expanding the determinant. A single sign error can lead to an incorrect result.
• Incorrect Operations: Ensure that operations are applied correctly. For example, $R_i \to R_i + kR_j$ changes only $R_i$, not $R_j$.
• Forgetting to Factor: After creating a common factor in a row/column, remember to actually factor it out of the determinant. This simplifies the remaining matrix.
• Not Utilizing Zeros: The primary goal of creating zeros is to simplify expansion. Always expand along the row or column containing the most zeros.

4. Summary

This problem demonstrates an effective strategy for evaluating determinants using row and column operations. The key steps involved transforming the determinant to reveal a common factor $(a+b+c)$, factoring it out, and then creating zeros in a row to simplify the final expansion. By systematically applying these properties, we reduced the determinant to $(a+b+c)^3$. Equating this with the given expression and solving for $x$ while respecting the condition $x \ne 0$ led us to the unique solution.

The final answer is $\boxed{\text{–2(a + b + c)}}$, which corresponds to option (A).