JEE Main 201911 Jan, Evening shiftMathematicsMediumMatrices and Determinants245× since 2002If ∣a−b−c2a2a2bb−c−a2b2c2cc−a−b∣\left| {\matrix{ {a - b - c} & {2a} & {2a} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right|a−b−c2b2c2ab−c−a2c2a2bc−a−b = (a + b + c) (x + a + b + c)², x ≠\ne= 0, then x is equal to :AOption A. –2(a + b + c)BOption B. 2(a + b + c)COption C. abcDOption D. –(a + b + c)CheckSkip · show solution
