Let the equation of a line through `(3, 6, -2)` and parallel to the planes `x-y+2z=5 and 3x+y+2z=6` is `L=0`. If point `(alpha,beta, 2)` satisfy `L=0`, then `alpha+2beta` is equal to

Options

A. 10
B. 13
C. 15
D. 19

Answer

To solve the problem step-by-step, we need to find the equation of the line that passes through the point (3, 6, -2) and is parallel to the given planes. Then, we will find the values of alpha and beta such that the point (alpha, beta, 2) lies on this line, and finally compute alpha + 2beta.

### Step 1: Find the normal vectors of the planes
The equations of the planes are:
1. \( x - y + 2z = 5 \)
2. \( 3x + y + 2z = 6 \)

The normal vector of the first plane is given by the coefficients of \(x\), \(y\), and \(z\):
- Normal vector \( \mathbf{n_1} = (1, -1, 2) \)

The normal vector of the second plane is:
- Normal vector \( \mathbf{n_2} = (3, 1, 2) \)

### Step 2: Find a direction vector for the line
The direction vector of the line that is parallel to both planes can be found using the cross product of the two normal vectors.

Let:
\[ \mathbf{n_1} = (1, -1, 2) \]
\[ \mathbf{n_2} = (3, 1, 2) \]

The cross product \( \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} \) is calculated as follows:
\[ \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ 3 & 1 & 2 \end{vmatrix} \]

Calculating the determinant:
\[ \mathbf{d} = \mathbf{i}((-1)(2) - (2)(1)) - \mathbf{j}((1)(2) - (2)(3)) + \mathbf{k}((1)(1) - (-1)(3)) \]
\[ = \mathbf{i}(-2 - 2) - \mathbf{j}(2 - 6) + \mathbf{k}(1 + 3) \]
\[ = -4\mathbf{i} + 4\mathbf{j} + 4\mathbf{k} \]
Thus, the direction vector \( \mathbf{d} = (-4, 4, 4) \).

### Step 3: Write the equation of the line
The line passing through the point \( (3, 6, -2) \) with direction vector \( (-4, 4, 4) \) can be expressed in parametric form:
\[ \frac{x - 3}{-4} = \frac{y - 6}{4} = \frac{z + 2}{4} \]

### Step 4: Substitute the point (alpha, beta, 2)
We need to find \( \alpha \) and \( \beta \) such that the point \( (\alpha, \beta, 2) \) lies on this line. Thus, we set:
\[ \frac{\alpha - 3}{-4} = \frac{\beta - 6}{4} = \frac{2 + 2}{4} \]

Calculating \( \frac{2 + 2}{4} = 1 \):
1. From \( \frac{\alpha - 3}{-4} = 1 \):
\[ \alpha - 3 = -4 \implies \alpha = -1 \]

2. From \( \frac{\beta - 6}{4} = 1 \):
\[ \beta - 6 = 4 \implies \beta = 10 \]

### Step 5: Calculate \( \alpha + 2\beta \)
Now we can find:
\[ \alpha + 2\beta = -1 + 2(10) = -1 + 20 = 19 \]

### Final Answer
Thus, \( \alpha + 2\beta = 19 \).

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