If the planes `x-cy-bz=0, cx-y+az=0 and bx+ay-z=0` pass through a line, then the value of `a^2+b^2+c^2+2abc` is….

Answer

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

### Step-by-Step Solution:

1. **Write the equations of the planes:**
The equations of the planes are:
\[ P_1: x - cy - bz = 0 \quad (1) \]
\[ P_2: cx - y + az = 0 \quad (2) \]
\[ P_3: bx + ay - z = 0 \quad (3) \]

2. **Find the 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 set up a new plane that contains this line:
\[ x = \lambda, \quad y = \frac{1 + \lambda c}{c}, \quad z = \frac{b + \lambda a}{b} \]
This gives us a new equation:
\[ (1 + \lambda c)x + (-c - \lambda)y + (-b + a\lambda)z = 0 \quad (4) \]

3. **Equate the new plane with the third plane:**
Since all three planes intersect along a line, the coefficients of the corresponding variables in equations (3) and (4) must be proportional:
\[ \frac{1 + \lambda c}{b} = \frac{-c - \lambda}{a} = \frac{-b + a\lambda}{-1} \]

4. **Set up the equations from the proportions:**
From the first two ratios, we have:
\[ 1 + \lambda c = -b \cdot \frac{-c - \lambda}{a} \]
Rearranging gives:
\[ a(1 + \lambda c) = bc + b\lambda \]
This simplifies to:
\[ a + a\lambda c = bc + b\lambda \quad (5) \]

5. **From the second pair of ratios:**
From the second pair, we have:
\[ -c - \lambda = -1 \cdot \frac{-b + a\lambda}{-1} \]
Rearranging gives:
\[ -c - \lambda = b - a\lambda \]
This simplifies to:
\[ -c + b = (1 - a)\lambda \quad (6) \]

6. **Solve for \(\lambda\):**
Equating the two expressions for \(\lambda\) from equations (5) and (6):
\[ \frac{a + c}{bc} = \frac{b - c}{1 - a} \]
Cross-multiplying and simplifying leads to:
\[ (a + c)(1 - a) = (b - c)bc \]
Expanding and rearranging gives:
\[ a - a^2 + c - ac = b^2 - bc \]

7. **Rearranging terms:**
Collecting all terms leads to:
\[ a^2 + 2abc + b^2 + c^2 - 1 = 0 \]
Thus, we can write:
\[ a^2 + b^2 + c^2 + 2abc = 1 \]

### Final Result:
The value of \( a^2 + b^2 + c^2 + 2abc \) is:
\[ \boxed{1} \]