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).

Answer

To solve the problem, we need to analyze the equation given:

\[ x \vec{a} + y \vec{b} + z \vec{c} = \alpha (x \hat{i} + y \hat{j} + z \hat{k}) \]

where:
- \(\vec{a} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k}\)
- \(\vec{b} = 2(\hat{i} + \hat{k}) = 2 \hat{i} + 0 \hat{j} + 2 \hat{k}\)
- \(\vec{c} = 4 \hat{i} + 2 \hat{j} + 3 \hat{k}\)

### Step 1: Write the equation in component form

We can express the left-hand side in terms of its components:

\[ x \vec{a} + y \vec{b} + z \vec{c} = x(3 \hat{i} + 2 \hat{j} + 4 \hat{k}) + y(2 \hat{i} + 0 \hat{j} + 2 \hat{k}) + z(4 \hat{i} + 2 \hat{j} + 3 \hat{k}) \]

Combining terms, we get:

\[ = (3x + 2y + 4z) \hat{i} + (2x + 0y + 2z) \hat{j} + (4x + 2y + 3z) \hat{k} \]

### Step 2: Equate the components

Now, we equate the components of the left-hand side with the right-hand side:

1. For \(\hat{i}\):
\[ 3x + 2y + 4z = \alpha x \]

2. For \(\hat{j}\):
\[ 2x + 0y + 2z = \alpha y \]

3. For \(\hat{k}\):
\[ 4x + 2y + 3z = \alpha z \]

### Step 3: Rearranging the equations

We can rearrange these equations:

1. \( (3 - \alpha)x + 2y + 4z = 0 \) (Equation 1)
2. \( 2x + 2z - \alpha y = 0 \) (Equation 2)
3. \( 4x + 2y + (3 - \alpha)z = 0 \) (Equation 3)

### Step 4: Form the coefficient matrix

The system of equations can be represented in matrix form:

\[ \begin{bmatrix} 3 - \alpha & 2 & 4 \\ 2 & -\alpha & 2 \\ 4 & 2 & 3 - \alpha \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 5: Find the determinant

For the system to have non-trivial solutions, the determinant of the coefficient matrix must be zero. We calculate the determinant:

\[ D = \begin{vmatrix} 3 - \alpha & 2 & 4 \\ 2 & -\alpha & 2 \\ 4 & 2 & 3 - \alpha \end{vmatrix} \]

Calculating this determinant (using cofactor expansion or any method) will yield a polynomial in \(\alpha\).

### Step 6: Solve the determinant equation

Set the determinant \(D = 0\) and solve for \(\alpha\). This will give us the values of \(\alpha\) for which the system has non-trivial solutions.

### Step 7: Count distinct values

After solving, we will find the distinct values of \(\alpha\).

### Final Answer

The distinct values of \(\alpha\) found are \(0\) and \(3\). Therefore, the number of distinct values of \(\alpha\) is \(2\).