The values of `k in R` for which the system of equations `x+k y+3z=0,k x+2y+2z=0,2x+3y+4z=0` admits of nontrivial solution is `2` b. `5//2` c. `3` d. `5//4`

Options

A. 2
B. `5//2`
C. `3`
D. `5//4`

Explanation

To find the values of \( k \in \mathbb{R} \) for which the system of equations

\[ \begin{align*} 1. & \quad x + ky + 3z = 0 \\ 2. & \quad kx + 2y + 2z = 0 \\ 3. & \quad 2x + 3y + 4z = 0 \end{align*} \]

admits a nontrivial solution, we need to set up the corresponding determinant of the coefficients of the variables \( x, y, z \) and find the values of \( k \) for which this determinant equals zero.

### Step 1: Write the coefficient matrix
The coefficient matrix \( A \) of the system is:

\[ A = \begin{bmatrix} 1 & k & 3 \\ k & 2 & 2 \\ 2 & 3 & 4 \end{bmatrix} \]

### Step 2: Calculate the determinant of the matrix
We need to calculate the determinant of matrix \( A \):

\[ \text{det}(A) = \begin{vmatrix} 1 & k & 3 \\ k & 2 & 2 \\ 2 & 3 & 4 \end{vmatrix} \]

Using the determinant formula for a \( 3 \times 3 \) matrix:

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 2 & 2 \\ 3 & 4 \end{vmatrix} - k \cdot \begin{vmatrix} k & 2 \\ 2 & 4 \end{vmatrix} + 3 \cdot \begin{vmatrix} k & 2 \\ 2 & 3 \end{vmatrix} \]

### Step 3: Calculate the minors
Now we calculate the minors:

1. \( \begin{vmatrix} 2 & 2 \\ 3 & 4 \end{vmatrix} = (2)(4) - (2)(3) = 8 - 6 = 2 \)

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

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

### Step 4: Substitute back into the determinant
Substituting these values back into the determinant expression gives:

\[ \text{det}(A) = 1 \cdot 2 - k(4k - 4) + 3(3k - 4) \]

This simplifies to:

\[ \text{det}(A) = 2 - (4k^2 - 4k) + (9k - 12) \]

Combining like terms:

\[ \text{det}(A) = 2 - 4k^2 + 4k + 9k - 12 = -4k^2 + 13k - 10 \]

### Step 5: Set the determinant to zero
For the system to have a nontrivial solution, we set the determinant equal to zero:

\[ -4k^2 + 13k - 10 = 0 \]

Multiplying through by -1 gives:

\[ 4k^2 - 13k + 10 = 0 \]

### Step 6: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:

\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \( a = 4, b = -13, c = 10 \):

\[ k = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 4 \cdot 10}}{2 \cdot 4} \]
\[ k = \frac{13 \pm \sqrt{169 - 160}}{8} \]
\[ k = \frac{13 \pm \sqrt{9}}{8} \]
\[ k = \frac{13 \pm 3}{8} \]

Calculating the two possible values:

1. \( k = \frac{16}{8} = 2 \)
2. \( k = \frac{10}{8} = \frac{5}{4} \)

### Final Answer
Thus, the values of \( k \) for which the system admits a nontrivial solution are \( k = 2 \) and \( k = \frac{5}{4} \).