JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Vector Algebra

Three vectors are coplanar when their scalar triple product, the determinant with the vectors as rows, is zero. That determinant for $\mu \hat{i}+\hat{j}+\hat{k}$, $\hat{i}+\mu \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+\mu \hat{k}$ factors neatly: adding all three rows into the first pulls out $(\mu +2)$, and the rest reduces to $(\mu -1{)}^2$, so the condition is $(\mu +2)(\mu -1{)}^2=0$. The distinct roots are $\mu =-2$ and $\mu =1$, and their sum is $-1$.