Given `vec(a)=3hati+2hatj+4hatk,vec(b)=2(hati+hatk)` and `vec( c )=4hati+2hatj+3hatk`. Find for what number of distinct values of `alpha` the equation `x vec(a)+y vec(b)+z vec( c )=alpha(x hati+y hatj+z hatk)` has non-trival solution (x, y, z).

Options

A. -1
B. 4
C. 7
D. 8

Answer

To solve the equation \( x \vec{a} + y \vec{b} + z \vec{c} = \alpha (x \hat{i} + y \hat{j} + z \hat{k}) \) for distinct values of \( \alpha \), we start by substituting the given vectors:

1. **Given Vectors:**
- \( \vec{a} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k} \)
- \( \vec{b} = 2 (\hat{i} + \hat{k}) = 2 \hat{i} + 0 \hat{j} + 2 \hat{k} \)
- \( \vec{c} = 4 \hat{i} + 2 \hat{j} + 3 \hat{k} \)

2. **Substituting into the Equation:**
\[ x(3 \hat{i} + 2 \hat{j} + 4 \hat{k}) + y(2 \hat{i} + 0 \hat{j} + 2 \hat{k}) + z(4 \hat{i} + 2 \hat{j} + 3 \hat{k}) = \alpha (x \hat{i} + y \hat{j} + z \hat{k}) \]

3. **Expanding the Left Side:**
\[ (3x + 2y + 4z) \hat{i} + (2x + 0y + 2z) \hat{j} + (4x + 2y + 3z) \hat{k} = \alpha (x \hat{i} + y \hat{j} + z \hat{k}) \]

4. **Equating Coefficients:**
- For \( \hat{i} \): \( 3x + 2y + 4z = \alpha x \)
- For \( \hat{j} \): \( 2x + 0y + 2z = \alpha y \)
- For \( \hat{k} \): \( 4x + 2y + 3z = \alpha z \)

5. **Rearranging the Equations:**
- \( (3 - \alpha)x + 2y + 4z = 0 \) (1)
- \( 2x + (0 - \alpha)y + 2z = 0 \) (2)
- \( 4x + 2y + (3 - \alpha)z = 0 \) (3)

6. **Setting up the Coefficient Matrix:**
The system can be represented as:
\[ \begin{bmatrix} 3 - \alpha & 2 & 4 \\ 2 & -\alpha & 2 \\ 4 & 2 & 3 - \alpha \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0 \]

7. **Finding the Determinant:**
For non-trivial solutions, the determinant of the coefficient matrix must equal zero:
\[ \text{Det} = \begin{vmatrix} 3 - \alpha & 2 & 4 \\ 2 & -\alpha & 2 \\ 4 & 2 & 3 - \alpha \end{vmatrix} = 0 \]

8. **Calculating the Determinant:**
Expanding the determinant:
\[ = (3 - \alpha)(-\alpha(3 - \alpha) - 4) - 2(2(3 - \alpha) - 8) + 4(4(-\alpha) - 8) \]
Simplifying this leads to a polynomial in \( \alpha \).

9. **Solving the Polynomial:**
After simplifying, we will find the roots of the polynomial which gives us the distinct values of \( \alpha \).

10. **Finding Distinct Values:**
The roots of the polynomial will provide the distinct values of \( \alpha \).

### Final Result:
After solving the polynomial, we find the distinct values of \( \alpha \) are \( -1 \) and \( 8 \). Therefore, there are **2 distinct values of \( \alpha \)**.