If `veca`,`vecb`,`vecc` are non-coplanar vectors and `lambda` is a real number then
[`lambda(veca+vecb)` `lambda^2vecb` `lambdavecc`] = [`veca` `vecb+vecc` `vecb`] for:

Options

A. (a)exactly two values of `lamda`
B. (b)exactly three values `lambda`
C. (c)no value of `lambda`
D. (d)exactly one value of `lambda`

Answer

To solve the problem, we need to analyze the equation given and simplify it step by step.

### Step 1: Write the Determinant
We are given the equation:

\[ \begin{vmatrix} \lambda(\vec{a} + \vec{b}) & \lambda^2 \vec{b} & \lambda \vec{c} \\ \end{vmatrix} = \begin{vmatrix} \vec{a} & \vec{b} + \vec{c} & \vec{b} \\ \end{vmatrix} \]

### Step 2: Expand the Left Side
The left side can be expanded using the properties of determinants. We can factor out the constants from the determinant:

\[ \lambda \cdot \lambda^2 \cdot \lambda = \lambda^4 \begin{vmatrix} \vec{a} + \vec{b} & \vec{b} & \vec{c} \\ \end{vmatrix} \]

### Step 3: Expand the Right Side
Now, we expand the right side. The determinant can be rewritten as:

\[ \begin{vmatrix} \vec{a} & \vec{b} + \vec{c} & \vec{b} \\ \end{vmatrix} \]

This can be simplified using the property of determinants that allows us to replace one column with the sum of two columns:

\[ = \begin{vmatrix} \vec{a} & \vec{b} & \vec{b} \\ \end{vmatrix} + \begin{vmatrix} \vec{a} & \vec{c} & \vec{b} \\ \end{vmatrix} \]

### Step 4: Set the Determinants Equal
Now we set the two sides equal:

\[ \lambda^4 \begin{vmatrix} \vec{a} + \vec{b} & \vec{b} & \vec{c} \\ \end{vmatrix} = \begin{vmatrix} \vec{a} & \vec{b} & \vec{b} \\ \end{vmatrix} + \begin{vmatrix} \vec{a} & \vec{c} & \vec{b} \\ \end{vmatrix} \]

### Step 5: Analyze the Determinants
Since \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar, the determinants will not be zero. Thus, we can analyze the equality of the two sides.

### Step 6: Solve for \(\lambda\)
From the equality, we can conclude that:

\[ \lambda^4 = 1 \]

This implies:

\[ \lambda = 1 \quad \text{or} \quad \lambda = -1 \]

### Conclusion
Thus, the values of \(\lambda\) that satisfy the equation are \(1\) and \(-1\).