If the vectors `veca, vecb, vecc` are non -coplanar and `l,m,n` are distinct scalars such that
`[(lveca+mvecb+nvecc, lvecb+mvecc+nveca,lvecc+mveca+nvecb)]=0` then

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 `

Explanation

To solve the problem step by step, we will analyze the given vectors and the condition provided.

### Step 1: Understand the Given Condition
We have three non-coplanar vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). The distinct scalars are \(l\), \(m\), and \(n\). The condition given is:

\[ \begin{vmatrix} l\vec{a} + m\vec{b} + n\vec{c} & l\vec{b} + m\vec{c} + n\vec{a} & l\vec{c} + m\vec{a} + n\vec{b} \end{vmatrix} = 0 \]

### Step 2: Set Up the Determinant
We can express the vectors in terms of their components. Let's denote:
- \(\vec{a} = (a_1, a_2, a_3)\)
- \(\vec{b} = (b_1, b_2, b_3)\)
- \(\vec{c} = (c_1, c_2, c_3)\)

The determinant can be expressed as:

\[ \begin{vmatrix} l a_1 + m b_1 + n c_1 & l b_1 + m c_1 + n a_1 & l c_1 + m a_1 + n b_1 \\ l a_2 + m b_2 + n c_2 & l b_2 + m c_2 + n a_2 & l c_2 + m a_2 + n b_2 \\ l a_3 + m b_3 + n c_3 & l b_3 + m c_3 + n a_3 & l c_3 + m a_3 + n b_3 \end{vmatrix} = 0 \]

### Step 3: Calculate the Determinant
We can denote the rows of the determinant as follows:
- First row: \((l a_1 + m b_1 + n c_1, l b_1 + m c_1 + n a_1, l c_1 + m a_1 + n b_1)\)
- Second row: \((l a_2 + m b_2 + n c_2, l b_2 + m c_2 + n a_2, l c_2 + m a_2 + n b_2)\)
- Third row: \((l a_3 + m b_3 + n c_3, l b_3 + m c_3 + n a_3, l c_3 + m a_3 + n b_3)\)

### Step 4: Apply Properties of Determinants
Since the determinant is equal to zero, it implies that the rows are linearly dependent. This means that there exist scalars (not all zero) such that:

\[ k_1 (l a_1 + m b_1 + n c_1) + k_2 (l b_1 + m c_1 + n a_1) + k_3 (l c_1 + m a_1 + n b_1) = 0 \]

### Step 5: Use the Identity for Cubes
From the determinant condition, we can derive that:

\[ l^3 + m^3 + n^3 - 3lmn = 0 \]

This is a known identity that holds true if \(l + m + n = 0\).

### Step 6: Conclusion
Thus, we conclude that:

\[ l + m + n = 0 \]

This means that the sum of the distinct scalars \(l\), \(m\), and \(n\) is equal to zero.