Question:

<p><strong>Step 1: Set up the determinant</strong></p><p>Let D = $\begin{vmatrix} a-b-c & 2a & 2a \ 2b & b-c-a & 2b \ 2c & 2c & c-a-b \end{vmatrix}$</p><p><strong>Step 2: Perform row operations (Add all rows to Row 1)</strong></p><p>R₁ → R₁ + R₂ + R₃</p><p>First row becomes:</p><p>$(a-b-c+2b+2c, 2a+b-c-a+2c, 2a+2b+c-a-b)$</p><p>$= (a+b+c, a+b+c, a+b+c)$</p><p>So D = $\begin{vmatrix} a+b+c & a+b+c & a+b+c \ 2b & b-c-a & 2b \ 2c & 2c & c-a-b \end{vmatrix}$</p><p><strong>Step 3: Factor out (a+b+c) from Row 1</strong></p><p>D = $(a+b+c)\begin{vmatrix} 1 & 1 & 1 \ 2b & b-c-a & 2b \ 2c & 2c & c-a-b \end{vmatrix}$</p><p><strong>Step 4: Perform column operations (C₂ → C₂ - C₁, C₃ → C₃ - C₁)</strong></p><p>D = $(a+b+c)\begin{vmatrix} 1 & 0 & 0 \ 2b & b-c-a-2b & 2b-2b \ 2c & 2c-2c & c-a-b-2c \end{vmatrix}$</p><p>$= (a+b+c)\begin{vmatrix} 1 & 0 & 0 \ 2b & -b-c-a & 0 \ 2c & 0 & -a-b-c \end{vmatrix}$</p><p><strong>Step 5: Expand along Row 1</strong></p><p>D = $(a+b+c)\cdot 1 \cdot \begin{vmatrix} -b-c-a & 0 \ 0 & -a-b-c \end{vmatrix}$</p><p>$= (a+b+c)(-b-c-a)(-a-b-c)$</p><p>$= (a+b+c) \cdot (-(a+b+c)) \cdot (-(a+b+c))$</p><p>$= (a+b+c)(a+b+c)^2$</p><p><strong>Step 6: Compare with given form</strong></p><p>Given: D = $(a+b+c)(x+a+b+c)^2$</p><p>From calculation: D = $(a+b+c)(a+b+c)^2$</p><p>Therefore: $(x+a+b+c)^2 = (a+b+c)^2$</p><p>This gives: $x + a + b + c = ±(a+b+c)$</p><p>If $x + a + b + c = a + b + c$, then $x = 0$ (excluded)</p><p>If $x + a + b + c = -(a+b+c)$, then $x = -2(a+b+c)$</p><p><strong>∴ Answer: b</strong></p>