If the planes `x=cy+bz,y=az+cx` and `z=bx+ay` pass through a line then `a^(2)+b^(2)+c^(2)+2abc` is equal to

Options

A. `-1`
B. 3
C. 1
D. 0

Answer

To solve the problem, we need to analyze the given planes and determine the condition under which they intersect along a line. The planes are given by the equations:

1. \( x = cy + bz \)
2. \( y = az + cx \)
3. \( z = bx + ay \)

### Step 1: Rewrite the equations in standard form
We can rewrite these equations in the standard form of a plane.

- From the first equation, we can rearrange it to:
\[ x - cy - bz = 0 \quad \text{(Plane 1)} \]

- From the second equation, we rearrange it to:
\[ cx - y + az = 0 \quad \text{(Plane 2)} \]

- From the third equation, we rearrange it to:
\[ bx + ay - z = 0 \quad \text{(Plane 3)} \]

### Step 2: Set up the system of equations
We now have a system of three equations:
1. \( x - cy - bz = 0 \)
2. \( cx - y + az = 0 \)
3. \( bx + ay - z = 0 \)

### Step 3: Formulate the coefficient matrix
The coefficient matrix \( A \) for this system is:
\[ A = \begin{bmatrix} 1 & -c & -b \\ c & -1 & a \\ b & a & -1 \end{bmatrix} \]

### Step 4: Find the determinant of the matrix
For the planes to intersect along a line, the determinant of the coefficient matrix must be zero:
\[ \text{det}(A) = 0 \]

Calculating the determinant:
\[ \text{det}(A) = 1 \cdot \begin{vmatrix} -1 & a \\ a & -1 \end{vmatrix} - (-c) \cdot \begin{vmatrix} c & a \\ b & -1 \end{vmatrix} - (-b) \cdot \begin{vmatrix} c & -1 \\ b & a \end{vmatrix} \]

Calculating the minors:
1. \( \begin{vmatrix} -1 & a \\ a & -1 \end{vmatrix} = (-1)(-1) - (a)(a) = 1 - a^2 \)

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

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

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

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

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

### Conclusion
Thus, the value of \( a^2 + b^2 + c^2 + 2abc \) is equal to \( 1 \).