Let a be a real number and `vec alpha = hati +2hatj, vec beta=2hati+a hatj+10 hatk, vec gamma=12hati+20hatj+a hatk` be three vectors, then `vec alpha, vec beta and vec gamma` are linearly independent for :

Options

A. `a gt 0`
B. `a lt 0`
C. `a=0`
D. No value of a

Answer

To determine the values of \( a \) for which the vectors \( \vec{\alpha}, \vec{\beta}, \) and \( \vec{\gamma} \) are linearly independent, we need to check the condition for linear independence using the determinant of the matrix formed by these vectors.

### Step 1: Write down the vectors
We have:
\[ \vec{\alpha} = \hat{i} + 2\hat{j} + 0\hat{k} = (1, 2, 0) \]
\[ \vec{\beta} = 2\hat{i} + a\hat{j} + 10\hat{k} = (2, a, 10) \]
\[ \vec{\gamma} = 12\hat{i} + 20\hat{j} + a\hat{k} = (12, 20, a) \]

### Step 2: Form the matrix
We form the matrix \( M \) using these vectors as rows:
\[ M = \begin{pmatrix} 1 & 2 & 0 \\ 2 & a & 10 \\ 12 & 20 & a \end{pmatrix} \]

### Step 3: Calculate the determinant of the matrix
To check for linear independence, we need to find the determinant of \( M \) and set it equal to zero:
\[ \text{det}(M) = \begin{vmatrix} 1 & 2 & 0 \\ 2 & a & 10 \\ 12 & 20 & a \end{vmatrix} \]

Using the formula for the determinant of a \( 3 \times 3 \) matrix:
\[ \text{det}(M) = 1 \cdot \begin{vmatrix} a & 10 \\ 20 & a \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 10 \\ 12 & a \end{vmatrix} + 0 \]

Calculating the first determinant:
\[ \begin{vmatrix} a & 10 \\ 20 & a \end{vmatrix} = a^2 - 200 \]

Calculating the second determinant:
\[ \begin{vmatrix} 2 & 10 \\ 12 & a \end{vmatrix} = 2a - 120 \]

Putting it all together:
\[ \text{det}(M) = 1(a^2 - 200) - 2(2a - 120) \]
\[ = a^2 - 200 - 4a + 240 \]
\[ = a^2 - 4a + 40 \]

### Step 4: Set the determinant to zero
To find the values of \( a \) for which the vectors are linearly independent, we set the determinant to zero:
\[ a^2 - 4a + 40 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -4, c = 40 \):
\[ a = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 40}}{2 \cdot 1} \]
\[ = \frac{4 \pm \sqrt{16 - 160}}{2} \]
\[ = \frac{4 \pm \sqrt{-144}}{2} \]
\[ = \frac{4 \pm 12i}{2} \]
\[ = 2 \pm 6i \]

Since the roots are complex, the determinant can never be zero for real values of \( a \). Therefore, the vectors are linearly independent for all real values of \( a \).

### Conclusion
The vectors \( \vec{\alpha}, \vec{\beta}, \) and \( \vec{\gamma} \) are linearly independent for all real numbers \( a \).