If the volume of a parallelopiped whose coterminus edges are a=i+j+2k, b=2i+`lambda`j+k and c=2i+2j+`lambda`k is 35 `m^3`, then a value of a.b + b.c - c.a is :

Options

A. `-10`
B. `2`
C. `22`
D. `-14`

Answer

To solve the problem, we need to find the value of \( a \cdot b + b \cdot c - c \cdot a \) given that the volume of the parallelepiped formed by the vectors \( a \), \( b \), and \( c \) is 35 \( m^3 \).

### Step 1: Write down the vectors
The vectors are given as:
- \( a = i + j + 2k \)
- \( b = 2i + \lambda j + k \)
- \( c = 2i + 2j + \lambda k \)

### Step 2: Find the volume of the parallelepiped
The volume \( V \) of the parallelepiped formed by the vectors \( a \), \( b \), and \( c \) is given by the absolute value of the scalar triple product, which can be computed using the determinant:

\[ V = |a \cdot (b \times c)| \]

This can be calculated using the determinant of a 3x3 matrix formed by the components of the vectors \( a \), \( b \), and \( c \):

\[ V = \left| \begin{vmatrix} 1 & 1 & 2 \\ 2 & \lambda & 1 \\ 2 & 2 & \lambda \end{vmatrix} \right| \]

### Step 3: Calculate the determinant
Calculating the determinant:

\[ \begin{vmatrix} 1 & 1 & 2 \\ 2 & \lambda & 1 \\ 2 & 2 & \lambda \end{vmatrix} = 1 \cdot \begin{vmatrix} \lambda & 1 \\ 2 & \lambda \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 1 \\ 2 & \lambda \end{vmatrix} + 2 \cdot \begin{vmatrix} 2 & \lambda \\ 2 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} \lambda & 1 \\ 2 & \lambda \end{vmatrix} = \lambda^2 - 2 \)

2. \( \begin{vmatrix} 2 & 1 \\ 2 & \lambda \end{vmatrix} = 2\lambda - 2 \)

3. \( \begin{vmatrix} 2 & \lambda \\ 2 & 2 \end{vmatrix} = 4 - 2\lambda \)

Putting it all together:

\[ = 1(\lambda^2 - 2) - 1(2\lambda - 2) + 2(4 - 2\lambda) \]

Simplifying:

\[ = \lambda^2 - 2 - 2\lambda + 2 + 8 - 4\lambda \]
\[ = \lambda^2 - 6\lambda + 8 \]

### Step 4: Set the volume equal to 35
Given that the volume is 35:

\[ |\lambda^2 - 6\lambda + 8| = 35 \]

This gives us two equations to solve:

1. \( \lambda^2 - 6\lambda + 8 = 35 \)
2. \( \lambda^2 - 6\lambda + 8 = -35 \)

### Step 5: Solve the first equation
\[ \lambda^2 - 6\lambda - 27 = 0 \]

Using the quadratic formula:

\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{36 + 108}}{2} = \frac{6 \pm \sqrt{144}}{2} = \frac{6 \pm 12}{2} \]

This gives:

\[ \lambda = 9 \quad \text{or} \quad \lambda = -3 \]

### Step 6: Solve the second equation
\[ \lambda^2 - 6\lambda + 43 = 0 \]

Calculating the discriminant:

\[ D = (-6)^2 - 4 \cdot 1 \cdot 43 = 36 - 172 = -136 \]

Since the discriminant is negative, this equation has no real solutions.

### Step 7: Calculate \( a \cdot b + b \cdot c - c \cdot a \)
Now we calculate \( a \cdot b + b \cdot c - c \cdot a \) for both values of \( \lambda \).

1. For \( \lambda = 9 \):
- \( a \cdot b = 2 + 9 + 2 = 13 \)
- \( b \cdot c = 4 + 18 + 9 = 31 \)
- \( c \cdot a = 4 + 2 + 18 = 24 \)

Therefore,
\[ a \cdot b + b \cdot c - c \cdot a = 13 + 31 - 24 = 20 \]

2. For \( \lambda = -3 \):
- \( a \cdot b = 2 - 3 + 2 = 1 \)
- \( b \cdot c = 4 - 6 - 3 = -5 \)
- \( c \cdot a = 4 + 2 - 6 = 0 \)

Therefore,
\[ a \cdot b + b \cdot c - c \cdot a = 1 - 5 - 0 = -4 \]

### Final Answer
The possible values of \( a \cdot b + b \cdot c - c \cdot a \) are \( 20 \) and \( -4 \). Since the problem asks for a value, we take \( 20 \).