The planes `x=cy+bz,y=az+cx,z=bx+ay` intersect in a line if `(a+b+c)^(2)` is equal to

Options

A. `1-abc+ab+ac+bc`
B. `a-2abc+2ab+2bc+2ca`
C. `1-abc+2ab+2bc+2ca`
D. none of these

Answer

To determine the condition under which the planes \( x = cy + bz \), \( y = az + cx \), and \( z = bx + ay \) intersect in a line, we can follow these steps:

### Step 1: Rewrite the equations of the planes
We start with the given equations of the planes:
1. \( x - cy - bz = 0 \) (Plane P1)
2. \( -cx + y - az = 0 \) (Plane P2)
3. \( -bx - ay + z = 0 \) (Plane P3)

### Step 2: Formulate the system of equations
We can represent the system of equations in matrix form:
\[ \begin{bmatrix} 1 & -c & -b \\ -c & 1 & -a \\ -b & -a & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 3: Determine the condition for a non-trivial solution
For the system to have a non-trivial solution (i.e., a solution other than \( x = 0, y = 0, z = 0 \)), the determinant of the coefficient matrix must be zero:
\[ \text{det} \begin{bmatrix} 1 & -c & -b \\ -c & 1 & -a \\ -b & -a & 1 \end{bmatrix} = 0 \]

### Step 4: Calculate the determinant
Calculating the determinant, we have:
\[ \text{det} = 1 \cdot \begin{vmatrix} 1 & -a \\ -a & 1 \end{vmatrix} - (-c) \cdot \begin{vmatrix} -c & -a \\ -b & 1 \end{vmatrix} - (-b) \cdot \begin{vmatrix} -c & 1 \\ -b & -a \end{vmatrix} \]

Calculating these 2x2 determinants:
1. \( \begin{vmatrix} 1 & -a \\ -a & 1 \end{vmatrix} = 1 + a^2 \)

2. \( \begin{vmatrix} -c & -a \\ -b & 1 \end{vmatrix} = -c + ab \)

3. \( \begin{vmatrix} -c & 1 \\ -b & -a \end{vmatrix} = ac - b \)

Putting it all together:
\[ \text{det} = 1(1 + a^2) + c(c - ab) + b(ac - b) \]
Simplifying this gives:
\[ 1 + a^2 + c^2 - abc + ac - b^2 \]

### Step 5: Set the determinant to zero
Setting the determinant to zero for the condition of intersection:
\[ 1 + a^2 + b^2 + c^2 - abc = 0 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ a^2 + b^2 + c^2 + 2abc = 1 \]

### Step 7: Find \( (a + b + c)^2 \)
Using the identity:
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \]
We substitute \( a^2 + b^2 + c^2 = 1 - 2abc \):
\[ (a + b + c)^2 = (1 - 2abc) + 2(ab + ac + bc) \]

### Conclusion
Thus, the condition for the planes to intersect in a line is:
\[ (a + b + c)^2 = 1 - 2abc + 2(ab + ac + bc) \]