If the points with position vectors `alphahati+10hatj+13hatk,6hati+11hatj+11hatk,9/2hati+betahatj-8hatk` are collinear, then `(19alpha-6beta)^(2)` is equal to .

Options

A. `36`
B. `16`
C. `49`
D. `25`

Answer

To solve the problem, we need to determine the value of \( (19\alpha - 6\beta)^2 \) given that the points with position vectors \( \vec{A} = \alpha \hat{i} + 10 \hat{j} + 13 \hat{k} \), \( \vec{B} = 6 \hat{i} + 11 \hat{j} + 11 \hat{k} \), and \( \vec{C} = \frac{9}{2} \hat{i} + \beta \hat{j} - 8 \hat{k} \) are collinear.

### Step 1: Find the vectors \( \vec{AB} \) and \( \vec{BC} \)

The vector \( \vec{AB} \) is given by:
\[ \vec{AB} = \vec{B} - \vec{A} = (6 - \alpha) \hat{i} + (11 - 10) \hat{j} + (11 - 13) \hat{k} = (6 - \alpha) \hat{i} + 1 \hat{j} - 2 \hat{k} \]

The vector \( \vec{BC} \) is given by:
\[ \vec{BC} = \vec{C} - \vec{B} = \left(\frac{9}{2} - 6\right) \hat{i} + (\beta - 11) \hat{j} + (-8 - 11) \hat{k} = \left(\frac{9}{2} - \frac{12}{2}\right) \hat{i} + (\beta - 11) \hat{j} - 19 \hat{k} \]
\[ = -\frac{3}{2} \hat{i} + (\beta - 11) \hat{j} - 19 \hat{k} \]

### Step 2: Set up the condition for collinearity

For the vectors \( \vec{AB} \) and \( \vec{BC} \) to be collinear, there must exist a scalar \( k \) such that:
\[ \vec{AB} = k \vec{BC} \]

This gives us the following equations:
1. \( 6 - \alpha = -\frac{3}{2}k \)
2. \( 1 = (\beta - 11)k \)
3. \( -2 = -19k \)

### Step 3: Solve for \( k \)

From the third equation:
\[ -2 = -19k \implies k = \frac{2}{19} \]

### Step 4: Substitute \( k \) back into the equations

Substituting \( k \) into the first equation:
\[ 6 - \alpha = -\frac{3}{2} \cdot \frac{2}{19} \implies 6 - \alpha = -\frac{3}{19} \implies \alpha = 6 + \frac{3}{19} = \frac{114}{19} \]

Substituting \( k \) into the second equation:
\[ 1 = (\beta - 11) \cdot \frac{2}{19} \implies \beta - 11 = \frac{19}{2} \implies \beta = 11 + \frac{19}{2} = \frac{41}{2} \]

### Step 5: Calculate \( 19\alpha - 6\beta \)

Now we can calculate \( 19\alpha - 6\beta \):
\[ 19\alpha = 19 \cdot \frac{114}{19} = 114 \]
\[ 6\beta = 6 \cdot \frac{41}{2} = 123 \]
\[ 19\alpha - 6\beta = 114 - 123 = -9 \]

### Step 6: Find \( (19\alpha - 6\beta)^2 \)

Finally, we calculate:
\[ (19\alpha - 6\beta)^2 = (-9)^2 = 81 \]

Thus, the value of \( (19\alpha - 6\beta)^2 \) is \( \boxed{81} \).