If the vectors $\vec a, \vec b,\vec c$ are non-coplanar and $l, m, n$ are distinct scalars such that $\begin{bmatrix}l\vec a+m\vec b +n\vec c& l\vec b+m\vec c+n\vec a& l\vec c +m\vec a+n\vec b\end{bmatrix}=0$ then
Answer & explanation
Correct answer: option 2
We have,
$\begin{bmatrix}l\vec a+m\vec b +n\vec c& l\vec b+m\vec c+n\vec a& l\vec c +m\vec a+n\vec b\end{bmatrix}=0$
$⇒\begin{bmatrix}l\vec a+m\vec b +n\vec c&n\vec a+l\vec b+m\vec c&m\vec a+n\vec b+l\vec c\end{bmatrix}$
$⇒\begin{vmatrix}l&m&n\\n&l&m\\m&m&l\end{vmatrix}\begin{bmatrix}\vec a&\vec b&\vec c\end{bmatrix}=0$
$⇒\begin{vmatrix}l&m&n\\n&l&m\\m&m&l\end{vmatrix}=0$
$⇒l^3+ m^3+n^3-3lmn=0$
$⇒(l+m+n) (l^2 + m^2 + n^2 - lm-mn-nl) = 0$
$⇒l+ m + n = 0$