Let `veca=2hati+3hatj-hatk` and `vecb=hati-2hatj+3hatk`. Then , the value of `lamda` for which the vector `vecc=lamdahati+hatj+(2lamda-1)hatk` is parallel to the plane containing `veca` and `vecb`. Is

Options

A. `1`
B. `0`
C. `-1`
D. `2`

Explanation

To solve the problem, we need to find the value of \( \lambda \) for which the vector \( \vec{c} = \lambda \hat{i} + \hat{j} + (2\lambda - 1) \hat{k} \) is parallel to the plane containing the vectors \( \vec{a} = 2\hat{i} + 3\hat{j} - \hat{k} \) and \( \vec{b} = \hat{i} - 2\hat{j} + 3\hat{k} \).

### Step 1: Understand the condition for parallelism
A vector \( \vec{c} \) is parallel to the plane formed by two vectors \( \vec{a} \) and \( \vec{b} \) if it is perpendicular to the cross product \( \vec{a} \times \vec{b} \). This means that \( \vec{c} \cdot (\vec{a} \times \vec{b}) = 0 \).

### Step 2: Calculate the cross product \( \vec{a} \times \vec{b} \)
We can calculate \( \vec{a} \times \vec{b} \) using the determinant of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of \( \vec{a} \) and \( \vec{b} \):

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & -2 & 3 \end{vmatrix} \]

### Step 3: Compute the determinant
Calculating the determinant, we have:

\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 3 & -1 \\ -2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -1 \\ 1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 1 & -2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ 3 \cdot 3 - (-1) \cdot (-2) = 9 - 2 = 7 \]

2. For \( \hat{j} \):
\[ 2 \cdot 3 - (-1) \cdot 1 = 6 + 1 = 7 \]

3. For \( \hat{k} \):
\[ 2 \cdot (-2) - 3 \cdot 1 = -4 - 3 = -7 \]

Putting it all together, we get:
\[ \vec{a} \times \vec{b} = 7\hat{i} - 7\hat{j} - 7\hat{k} = 7(\hat{i} - \hat{j} - \hat{k}) \]

### Step 4: Set up the dot product equation
Now, we need to find \( \lambda \) such that:
\[ \vec{c} \cdot (\vec{a} \times \vec{b}) = 0 \]

Substituting \( \vec{c} \) and \( \vec{a} \times \vec{b} \):
\[ (\lambda \hat{i} + \hat{j} + (2\lambda - 1) \hat{k}) \cdot (7\hat{i} - 7\hat{j} - 7\hat{k}) = 0 \]

Calculating the dot product:
\[ 7\lambda - 7 + (-7)(2\lambda - 1) = 0 \]

Expanding this:
\[ 7\lambda - 7 - 14\lambda + 7 = 0 \]

Combining like terms:
\[ -7\lambda = 0 \]

### Step 5: Solve for \( \lambda \)
Thus, we find:
\[ \lambda = 0 \]

### Final Answer
The value of \( \lambda \) for which the vector \( \vec{c} \) is parallel to the plane containing \( \vec{a} \) and \( \vec{b} \) is \( \lambda = 0 \).

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