JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Determinants

Adding the second and third rows to the first makes every entry of the new top row equal to $a+b+c$, so $(a+b+c)$ factors out and the first row becomes $1,1,1$. Subtracting the first column from each of the other two then reduces the top row to $1,0,0$ and leaves a diagonal lower block whose two non-zero entries are $-(a+b+c)$, so the determinant equals $(a+b+c)\cdot (a+b+c{)}^2=(a+b+c{)}^3$. Setting this equal to $(a+b+c)(x+a+b+c{)}^2$ and cancelling the non-zero factor $a+b+c$ gives $(x+a+b+c{)}^2=(a+b+c{)}^2$, hence $x+a+b+c=\pm (a+b+c)$. The positive sign forces $x=0$, which is ruled out, leaving $x=-2(a+b+c)$.