If the vectors `vec a, vec b, vec c` are non-coplanar and l,m,n are distinct real numbers, then [(`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`)] = 0, implies (A) `lm+mn+nl = 0` (B) `l+m+n = 0` (C) `l^2 + m^2 + n^2 = 0`

Options

A. ` l + m + n=0`
B. roots of the equation `lx^(2) + mx + n =0` are equal
C. `l^(2)+m^(2) + n^(2) =0`
D. `l^(3) + m^(2) + n^(3) = 3 lmn `

Answer

a,b,d

Explanation

`{:(vecV_(1)=lveca+mvecb+nvecc),(vecV_(2)=n veca+lvecb + mvecc),(vecV_(3) = mveca + nvecb+lvecc):}}when,veca, vecb and vecc " are non- coplanar".`
Therefore,
`[vecV_(1)vecV_(2)vecV_(3)]= |{:(l,m,n),(n,l,m),(m,n,l):}|=0`
`or (l + m +n) [(l-m)^(2) + (m-n)^(2)+ (n-l)^(2)=0`
` or l + m+n =0`
obviously, `lx^(2) + mx + n =0` is satisfied by x =1 due to (i).
` l^(3) +m^(3) + n^(3) = 3lmn `
`Rightarrow (l+m +n) (l^(2) + m^(2) +n^(2)-lm -mn -ln) =0`
which is true