If planes `x-cy-bz=0,cx-y+az=0` and `bx+ay-z=0` pass through a straight line then `a^(2)+b^(2)+c^(2)=`

Options

A. `1-abc`
B. `abc-1`
C. `1-2abc`
D. `2abc-1`

Answer

To solve the problem, we need to determine the value of \( a^2 + b^2 + c^2 \) given that the three planes \( x - cy - bz = 0 \), \( cx - y + az = 0 \), and \( bx + ay - z = 0 \) pass through a straight line.

### Step-by-Step Solution:

1. **Identify the equations of the planes**:
The equations of the planes are:
\[ P_1: x - cy - bz = 0 \]
\[ P_2: cx - y + az = 0 \]
\[ P_3: bx + ay - z = 0 \]

2. **Check if the planes pass through the origin**:
To check if the planes pass through the origin, we substitute \( x = 0 \), \( y = 0 \), and \( z = 0 \) into each plane equation:
- For \( P_1 \): \( 0 - c(0) - b(0) = 0 \) (True)
- For \( P_2 \): \( c(0) - 0 + a(0) = 0 \) (True)
- For \( P_3 \): \( b(0) + a(0) - 0 = 0 \) (True)

Since all three equations are satisfied, the planes pass through the origin.

3. **Set up the determinant condition**:
For three planes to intersect along a line, the determinant of their coefficients must be zero. We write the coefficient matrix:
\[ \begin{vmatrix} 1 & -c & -b \\ c & -1 & a \\ b & a & -1 \end{vmatrix} = 0 \]

4. **Calculate the determinant**:
We will expand the determinant:
\[ \text{Det} = 1 \cdot \begin{vmatrix} -1 & a \\ a & -1 \end{vmatrix} - (-c) \cdot \begin{vmatrix} c & a \\ b & -1 \end{vmatrix} - b \cdot \begin{vmatrix} c & -1 \\ b & a \end{vmatrix} \]

Calculating each of these 2x2 determinants:
- \( \begin{vmatrix} -1 & a \\ a & -1 \end{vmatrix} = (-1)(-1) - (a)(a) = 1 - a^2 \)
- \( \begin{vmatrix} c & a \\ b & -1 \end{vmatrix} = (c)(-1) - (a)(b) = -c - ab \)
- \( \begin{vmatrix} c & -1 \\ b & a \end{vmatrix} = (c)(a) - (-1)(b) = ac + b \)

Substituting these back into the determinant:
\[ \text{Det} = 1 - a^2 + c(c + ab) - b(ac + b) = 0 \]

5. **Simplify the determinant equation**:
Expanding and simplifying:
\[ 1 - a^2 + c^2 + abc - abc - b^2 = 0 \]
This simplifies to:
\[ 1 - a^2 - b^2 - c^2 = 0 \]

6. **Rearranging the equation**:
We can rearrange this equation to find \( a^2 + b^2 + c^2 \):
\[ a^2 + b^2 + c^2 = 1 - 2abc \]

### Final Answer:
Thus, the value of \( a^2 + b^2 + c^2 \) is:
\[ \boxed{1 - 2abc} \]