Correct Answer - Option 3 : (a + b + c)
3
Given:
\(\left| {\begin{array}{*{20}{c}} {a - b - c}&{2a}&{2a}\\ {2b}&{b - c - a}&{2b}\\ {2c}&{2c}&{c - a - b} \end{array}} \right|\)
Formula:
\(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]\)
|A| = a1(a22 × a33 - a32 × a23) - a12 (a21 × a33 - a31 × a23) + a13(a21 × a32 - a31 × a22)
Calculation:
\(Let \;A = \left[ {\begin{array}{*{20}{c}} {a - b - c}&{2a}&{2a}\\ {2b}&{b - c - a}&{2b}\\ {2c}&{2c}&{c - a - b} \end{array}} \right]\)
C1 → C1 - C2
\(A = \left[ {\begin{array}{*{20}{c}} {-(a + b + c)}&{2a}&{2a}\\ {a+b+c}&{b - c - a}&{2b}\\ {0}&{2c}&{c - a - b} \end{array}} \right]\)
|A| = -(a + b + c) [(b - c - a) (c - a - b) - 2b.2c)] -(a + b + c) [2a (c - a - b) - 2a. 2c] +0
= -(a + b + c) [(b - c - a) (c - a - b) - 4bc) + (2a (c - a - b) - 4ac]]
= -(a + b + c) [(b - c - a) (c - a - b) - 4bc + 2a (c - a - b) - 4ac]
= -(a + b + c) [bc - ab - b2 - c2 + ac + bc - ac + a2 + ab - 4bc + 2ac - 2ab - 2a2 - 4ac]
= -(a + b + c) [-a2 - b2 - c2 + 2bc - 4bc + 2ac - 4ac - 2ab]
= -(a + b + c) [ -a2 - b2 - c2 - 2bc - 2ac - 2ab]
= -(a + b + c) [ - (a2 + b2 + c2 + 2ab + 2bc + 2ac)]
= (a + b + c) (a2 + b2 + c2 + 2ab + 2bc + 2ac)
= (a + b + c)3
|A| = (a + b + c)3