Let `vec(x),vec(y)` be two non-collinear vectors such that.
`(l+2m)vec(x)+(l+2n)vec(y)=0` and `A(vec(a)),B(vec(b)),C(vec( c ))` are three points satisfying the relation `l vec(a)+m vec(b)+n vec( c )=0`. For any point P in space `(bar(PA)xx bar(PB))xx[(bar(PB)xx bar(PC))xx(bar(PC)xx bar(PA))]` is equal to

Options

A. `[bar(PA)bar(PB)bar(PC)]bar(PA)+bar(PB)`
B. `bar(PA)xx bar(PB)`
C. `bar(PB)xx bar(PC)`
D. `bar(0)`

Answer

To solve the problem, we need to analyze the given equations and the relationships between the vectors involved.

### Step-by-Step Solution:

1. **Understanding the Given Equations**:
We are given two non-collinear vectors \( \vec{x} \) and \( \vec{y} \) such that:
\[ (l + 2m)\vec{x} + (l + 2n)\vec{y} = 0 \]
This implies that both coefficients must be zero for the equation to hold (since \( \vec{x} \) and \( \vec{y} \) are non-collinear). Thus, we have:
\[ l + 2m = 0 \quad \text{(1)} \]
\[ l + 2n = 0 \quad \text{(2)} \]

2. **Solving for \( m \) and \( n \)**:
From equation (1), we can express \( m \) in terms of \( l \):
\[ m = -\frac{l}{2} \]
From equation (2), we can express \( n \) in terms of \( l \):
\[ n = -\frac{l}{2} \]

3. **Using the Position Vectors**:
We are also given the relation:
\[ l\vec{a} + m\vec{b} + n\vec{c} = 0 \]
Substituting the values of \( m \) and \( n \):
\[ l\vec{a} - \frac{l}{2}\vec{b} - \frac{l}{2}\vec{c} = 0 \]
Factoring out \( l \) (assuming \( l \neq 0 \)):
\[ \vec{a} - \frac{1}{2}\vec{b} - \frac{1}{2}\vec{c} = 0 \]
Rearranging gives:
\[ 2\vec{a} = \vec{b} + \vec{c} \]
This shows that \( \vec{a}, \vec{b}, \vec{c} \) are collinear.

4. **Finding the Vector Expression**:
We need to evaluate:
\[ (\bar{PA} \times \bar{PB}) \times [(\bar{PB} \times \bar{PC}) \times (\bar{PC} \times \bar{PA})] \]
To simplify the calculations, we can set the position vector of point \( P \) as \( \vec{0} \).

5. **Substituting \( P \) as the Origin**:
Let \( \bar{PA} = \vec{a} \), \( \bar{PB} = \vec{b} \), and \( \bar{PC} = \vec{c} \). Therefore, we need to compute:
\[ (\vec{a} \times \vec{b}) \times [(\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a})] \]

6. **Using the Vector Triple Product Identity**:
Using the identity \( \vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w} \):
- Let \( \vec{u} = \vec{a} \), \( \vec{v} = \vec{b} \), \( \vec{w} = \vec{c} \).
- First calculate \( (\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a}) \).

7. **Final Calculation**:
After performing the calculations, we find that the expression evaluates to zero because \( \vec{a}, \vec{b}, \vec{c} \) are collinear, leading to:
\[ \text{Result} = 0 \]