For real numbers `alpha and beta ne 0 ` , if the point of intersection of the straight lines ` ( x - alpha)/( 1) = ( y - 1)/( 2) = ( z -1)/( 3) and ( x - 4) / ( beta) = ( y - 6)/( 3) = ( z - 7)/(3)` , lies on the plane ` x + 2y - z = 8 ` then ` alpha - beta ` is equal to

Options

A. 5
B. 9
C. 3
D. 7

Answer

To solve the problem step by step, we need to find the point of intersection of the two lines and then check if that point lies on the given plane. Finally, we will calculate \( \alpha - \beta \).

### Step 1: Parametrize the lines
The first line \( L_1 \) is given by:
\[ \frac{x - \alpha}{1} = \frac{y - 1}{2} = \frac{z - 1}{3} = t \]
From this, we can express the coordinates in terms of the parameter \( t \):
\[ x = \alpha + t, \quad y = 1 + 2t, \quad z = 1 + 3t \]

The second line \( L_2 \) is given by:
\[ \frac{x - 4}{\beta} = \frac{y - 6}{3} = \frac{z - 7}{3} = s \]
From this, we can express the coordinates in terms of the parameter \( s \):
\[ x = 4 + \beta s, \quad y = 6 + 3s, \quad z = 7 + 3s \]

### Step 2: Set the coordinates equal to find the intersection
To find the point of intersection, we set the coordinates from both lines equal to each other:
1. \( \alpha + t = 4 + \beta s \)
2. \( 1 + 2t = 6 + 3s \)
3. \( 1 + 3t = 7 + 3s \)

### Step 3: Solve the equations
From the second equation:
\[ 1 + 2t = 6 + 3s \implies 2t - 3s = 5 \quad \text{(Equation 1)} \]

From the third equation:
\[ 1 + 3t = 7 + 3s \implies 3t - 3s = 6 \implies t - s = 2 \quad \text{(Equation 2)} \]

Now we have two equations:
1. \( 2t - 3s = 5 \)
2. \( t - s = 2 \)

### Step 4: Substitute to find \( t \) and \( s \)
From Equation 2, we can express \( t \) in terms of \( s \):
\[ t = s + 2 \]

Substituting this into Equation 1:
\[ 2(s + 2) - 3s = 5 \implies 2s + 4 - 3s = 5 \implies -s + 4 = 5 \implies -s = 1 \implies s = -1 \]

Now substituting back to find \( t \):
\[ t = -1 + 2 = 1 \]

### Step 5: Find the coordinates of the intersection point
Now substituting \( t = 1 \) into the parametric equations for \( L_1 \):
\[ x = \alpha + 1, \quad y = 1 + 2(1) = 3, \quad z = 1 + 3(1) = 4 \]
Thus, the point of intersection is:
\[ (\alpha + 1, 3, 4) \]

### Step 6: Substitute into the plane equation
Now, we need to check if this point lies on the plane \( x + 2y - z = 8 \):
\[ (\alpha + 1) + 2(3) - 4 = 8 \]
Simplifying:
\[ \alpha + 1 + 6 - 4 = 8 \implies \alpha + 3 = 8 \implies \alpha = 5 \]

### Step 7: Find \( \beta \)
Using the equation \( \alpha + \beta = 3 \):
\[ 5 + \beta = 3 \implies \beta = 3 - 5 = -2 \]

### Step 8: Calculate \( \alpha - \beta \)
Now we can find \( \alpha - \beta \):
\[ \alpha - \beta = 5 - (-2) = 5 + 2 = 7 \]

Thus, the final answer is:
\[ \alpha - \beta = 7 \]