If the planes `x-c y-b z=0,c x=y+a z=0a n db x+a y-z=0` pass through a straight line, then find the value of `a^2+b^2+c^2+2a b cdot`

Explanation

To solve the problem, we need to find the value of \( a^2 + b^2 + c^2 + 2ab \) given that the three planes intersect along a straight line. The equations of the planes are:

1. \( x - cy - bz = 0 \) (Let's call this Plane 1)
2. \( cx - y + az = 0 \) (Let's call this Plane 2)
3. \( bx + ay - z = 0 \) (Let's call this Plane 3)

### Step 1: Write the equations of the planes in standard form
We can rewrite the equations of the planes as follows:

1. \( x - cy - bz = 0 \) → \( x - cy - bz = 0 \)
2. \( cx - y + az = 0 \) → \( cx - y + az = 0 \)
3. \( bx + ay - z = 0 \) → \( bx + ay - z = 0 \)

### Step 2: Find the line of intersection of the first two planes
To find the line of intersection of the first two planes, we can express the equations in a parametric form. We can take the first plane and add a parameter \( \lambda \) times the second plane:

\[ P_1 + \lambda P_2 = 0 \]

This gives:

\[ (x - cy - bz) + \lambda(cx - y + az) = 0 \]

### Step 3: Combine the equations
Expanding this, we have:

\[ x - cy - bz + \lambda(cx - y + az) = 0 \]

This can be rearranged to:

\[ (1 + \lambda c)x + (-c + \lambda)y + (-b + \lambda a)z = 0 \]

### Step 4: Set the third plane equal to the combined equation
Since the third plane must also intersect along this line, we set the coefficients of \( x, y, z \) from the combined equation equal to those from the third plane:

\[ bx + ay - z = 0 \]

### Step 5: Set up the system of equations
From the coefficients, we can set up the following equations:

1. \( 1 + \lambda c = b \)
2. \( -c + \lambda = a \)
3. \( -b + \lambda a = -1 \)

### Step 6: Solve the system of equations
From the first equation, we can express \( \lambda \):

\[ \lambda = \frac{b - 1}{c} \]

Substituting this into the second equation gives:

\[ -c + \frac{b - 1}{c} = a \]

This leads to:

\[ -c + \frac{b - 1}{c} = a \implies -c^2 + b - 1 = ac \]

From the third equation, we can substitute \( \lambda \):

\[ -b + \frac{b - 1}{c}a = -1 \]

This gives us a system of equations that we can solve for \( a, b, c \).

### Step 7: Find the determinant condition
For the planes to intersect along a line, the determinant of the coefficients must be zero:

\[ \begin{vmatrix} 1 & -c & -b \\ c & -1 & a \\ b & a & -1 \end{vmatrix} = 0 \]

### Step 8: Calculate the determinant
Calculating the determinant, we find:

\[ 1(-1)(-1) + c(a)(b) + b(c)(-1) - (-b)(-1)(b) - (-c)(a)(1) - (1)(c)(a) = 0 \]

This simplifies to:

\[ 1 + abc - b^2 - ac - ab = 0 \]

### Step 9: Rearranging the equation
Rearranging gives us:

\[ a^2 + b^2 + c^2 + 2ab = 1 \]

### Final Step: Conclusion
Thus, we find that:

\[ a^2 + b^2 + c^2 + 2ab = 1 \]