JEE Main · 2009 · 12 Apr 2009 · Offline · Determinants

Rearranging the columns of the second determinant and pulling out the sign factors turns it into $(-1{)}^{n+2}$ times the first determinant. The sum of the two determinants is then $\left[1+(-1{)}^{n+2}\right]$ times a determinant that is not itself zero because $b(a+c)\ne 0$. The whole expression therefore vanishes only when $1+(-1{)}^{n+2}=0$, that is when $(-1{)}^{n+2}=-1$, which needs $n+2$ odd, so $n$ is any odd integer.