Let the shortest distance between the lines `L:(x-5)/(-2)=(y-lambda)/(0)=(z+lambda)/(1),lambdage0` and `L1:x+1=y-1=4-z` be `2sqrt(6)`.If `(alpha,beta,gamma)` lies on `L`,then which of the following is NOT possible?

Options

A. `alpha+2 beta=24`
B. `2 alpha+gamma=7`
C. `alpha-2 gamma=19`
D. `2 alpha-gamma=9`

Answer

To solve the problem, we need to find the shortest distance between the two lines \( L \) and \( L_1 \) and determine which of the given options is not possible for the coordinates \( (\alpha, \beta, \gamma) \) that lie on line \( L \).

### Step 1: Identify the direction ratios and points on the lines

For line \( L \):
\[ \frac{x - 5}{-2} = \frac{y - \lambda}{0} = \frac{z + \lambda}{1} \]
The direction ratios of line \( L \) are \( (-2, 0, 1) \).

For line \( L_1 \):
\[ x + 1 = y - 1 = 4 - z \]
The direction ratios of line \( L_1 \) are \( (1, 1, -1) \).

### Step 2: Calculate the cross product of the direction ratios

Let \( \mathbf{b_1} = (-2, 0, 1) \) and \( \mathbf{b_2} = (1, 1, -1) \).

The cross product \( \mathbf{b_1} \times \mathbf{b_2} \) is calculated as follows:
\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 0 & 1 \\ 1 & 1 & -1 \end{vmatrix} \]
Calculating the determinant:
\[ = \mathbf{i} \left(0 \cdot (-1) - 1 \cdot 1\right) - \mathbf{j} \left(-2 \cdot (-1) - 1 \cdot 1\right) + \mathbf{k} \left(-2 \cdot 1 - 0 \cdot 1\right) \]
\[ = -\mathbf{i} - (2 - 1)\mathbf{j} - 2\mathbf{k} = -\mathbf{i} - \mathbf{j} - 2\mathbf{k} \]
Thus, \( \mathbf{b_1} \times \mathbf{b_2} = (-1, -1, -2) \).

### Step 3: Find the vector between points on the two lines

Let \( A_1 = (5, \lambda, -\lambda) \) be a point on line \( L \) and \( A_2 = (-1, 1, 4) \) be a point on line \( L_1 \).

The vector \( \mathbf{A_2 - A_1} \) is:
\[ \mathbf{A_2 - A_1} = (-1 - 5, 1 - \lambda, 4 + \lambda) = (-6, 1 - \lambda, 4 + \lambda) \]

### Step 4: Use the formula for the shortest distance between skew lines

The formula for the shortest distance \( d \) between two skew lines is given by:
\[ d = \frac{|(\mathbf{A_2 - A_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]

Calculating the dot product:
\[ \mathbf{A_2 - A_1} \cdot (\mathbf{b_1} \times \mathbf{b_2}) = (-6, 1 - \lambda, 4 + \lambda) \cdot (-1, -1, -2) \]
\[ = 6 - (1 - \lambda) \cdot 1 - 2(4 + \lambda) = 6 - (1 - \lambda) - (8 + 2\lambda) = 6 - 1 + \lambda - 8 - 2\lambda = -3 - \lambda \]

Calculating the magnitude of the cross product:
\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{(-1)^2 + (-1)^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]

Thus, the distance \( d \) is:
\[ d = \frac{| -3 - \lambda |}{\sqrt{6}} \]

### Step 5: Set the distance equal to \( 2\sqrt{6} \)

Setting the distance equal to \( 2\sqrt{6} \):
\[ \frac{| -3 - \lambda |}{\sqrt{6}} = 2\sqrt{6} \]
Multiplying both sides by \( \sqrt{6} \):
\[ | -3 - \lambda | = 12 \]

This gives us two cases:
1. \( -3 - \lambda = 12 \) → \( \lambda = -15 \)
2. \( -3 - \lambda = -12 \) → \( \lambda = 9 \)

Since \( \lambda \geq 0 \), we take \( \lambda = 9 \).

### Step 6: Determine possible values for \( \alpha, \beta, \gamma \)

From the line \( L \):
\[ \alpha = -2k + 5, \quad \beta = \lambda = 9, \quad \gamma = k - 9 \]

Now, we can express \( \alpha + 2\gamma \):
\[ \alpha + 2\gamma = (-2k + 5) + 2(k - 9) = -2k + 5 + 2k - 18 = -13 \]

### Step 7: Check the options

We need to determine which of the following is NOT possible:
1. \( \alpha + 2\beta = 24 \)
2. \( \alpha + 2\gamma = -13 \)
3. \( \alpha + 2\beta = 35 \)

From our calculations:
- \( \alpha + 2\gamma = -13 \) is possible.
- \( \alpha + 2\beta = 24 \) gives \( -2k + 5 + 18 = 24 \) → \( -2k + 23 = 24 \) → \( k = -\frac{1}{2} \) (possible).
- \( \alpha + 2\beta = 35 \) gives \( -2k + 5 + 18 = 35 \) → \( -2k + 23 = 35 \) → \( -2k = 12 \) → \( k = -6 \) (possible).

Thus, the option that is NOT possible is:
\[ \alpha + 2\beta = 35 \]

### Final Answer
The option that is NOT possible is \( \alpha + 2\beta = 35 \).