For what value of `k` the following system of linear equations wil have infinite solutions `x-y+z=3, 2x+y-z=2` and `-3x-2ky+6z=3`

Options

A. `k!=2`
B. `k=0`
C. `k=3`
D. `k epsilon[2,3]`

Explanation

To find the value of \( k \) for which the given system of linear equations has infinite solutions, we can follow these steps:

### Given Equations:
1. \( x - y + z = 3 \) (Equation 1)
2. \( 2x + y - z = 2 \) (Equation 2)
3. \( -3x - 2ky + 6z = 3 \) (Equation 3)

### Step 1: Write the system in matrix form
We can represent the system of equations in the form of a matrix \( A \) and a vector \( B \):
\[ A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -3 & -2k & 6 \end{bmatrix}, \quad B = \begin{bmatrix} 3 \\ 2 \\ 3 \end{bmatrix} \]

### Step 2: Find the determinant of matrix \( A \)
To determine the value of \( k \) for which the system has infinite solutions, we need to find the determinant of matrix \( A \) and set it equal to zero:
\[ \text{det}(A) = \begin{vmatrix} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -3 & -2k & 6 \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the formula for the determinant of a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the remaining rows.

Calculating:
\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 1 & -1 \\ -2k & 6 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 2 & -1 \\ -3 & 6 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 1 \\ -3 & -2k \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 1 & -1 \\ -2k & 6 \end{vmatrix} = (1)(6) - (-1)(-2k) = 6 - 2k \)

2. \( \begin{vmatrix} 2 & -1 \\ -3 & 6 \end{vmatrix} = (2)(6) - (-1)(-3) = 12 - 3 = 9 \)

3. \( \begin{vmatrix} 2 & 1 \\ -3 & -2k \end{vmatrix} = (2)(-2k) - (1)(-3) = -4k + 3 \)

Now substituting back into the determinant:
\[ \text{det}(A) = 1(6 - 2k) + 9 + 1(-4k + 3) \]
\[ = 6 - 2k + 9 - 4k + 3 \]
\[ = 18 - 6k \]

### Step 4: Set the determinant equal to zero
For the system to have infinite solutions:
\[ 18 - 6k = 0 \]

### Step 5: Solve for \( k \)
\[ -6k = -18 \implies k = \frac{-18}{-6} = 3 \]

### Conclusion
The value of \( k \) for which the system of equations has infinite solutions is:
\[ \boxed{3} \]