The values of k for which lines `kx + 2y + 2 = 0, 2x + ky + 3 = 0, 3x + 3y + k = 0` are concurrent are

Options

A. `{2,3,5}`
B. `{2,3,-5}`
C. `{3,-5}`
D. `{-5}`

Explanation

To determine the values of \( k \) for which the lines \( kx + 2y + 2 = 0 \), \( 2x + ky + 3 = 0 \), and \( 3x + 3y + k = 0 \) are concurrent, we will follow these steps:

### Step 1: Set up the determinant for concurrency
The lines are concurrent if the determinant of their coefficients is zero. The coefficients of the lines can be arranged in a matrix as follows:

\[ \begin{vmatrix} k & 2 & 2 \\ 2 & k & 3 \\ 3 & 3 & k \end{vmatrix} = 0 \]

### Step 2: Calculate the determinant
We will expand the determinant:

\[ D = k \begin{vmatrix} k & 3 \\ 3 & k \end{vmatrix} - 2 \begin{vmatrix} 2 & 3 \\ 3 & k \end{vmatrix} + 2 \begin{vmatrix} 2 & k \\ 3 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

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

2. \( \begin{vmatrix} 2 & 3 \\ 3 & k \end{vmatrix} = 2k - 9 \)

3. \( \begin{vmatrix} 2 & k \\ 3 & 3 \end{vmatrix} = 6 - 3k \)

Now substituting these back into the determinant \( D \):

\[ D = k(k^2 - 9) - 2(2k - 9) + 2(6 - 3k) \]

### Step 3: Simplify the determinant
Expanding this gives:

\[ D = k^3 - 9k - 4k + 18 + 12 - 6k \]
\[ D = k^3 - 19k + 30 \]

### Step 4: Set the determinant to zero
To find the values of \( k \), we set the determinant equal to zero:

\[ k^3 - 19k + 30 = 0 \]

### Step 5: Find the roots of the cubic equation
We can use the Rational Root Theorem to test possible rational roots. Testing \( k = 2 \):

\[ 2^3 - 19(2) + 30 = 8 - 38 + 30 = 0 \]

Thus, \( k = 2 \) is a root. Now we can factor the cubic polynomial by dividing it by \( k - 2 \):

Using synthetic division or polynomial long division, we find:

\[ k^3 - 19k + 30 = (k - 2)(k^2 + 2k - 15) \]

### Step 6: Factor the quadratic
Next, we factor \( k^2 + 2k - 15 \):

\[ k^2 + 2k - 15 = (k - 3)(k + 5) \]

### Step 7: Combine the factors
Thus, the complete factorization of the cubic equation is:

\[ (k - 2)(k - 3)(k + 5) = 0 \]

### Step 8: Solve for \( k \)
Setting each factor to zero gives us the possible values of \( k \):

1. \( k - 2 = 0 \) → \( k = 2 \)
2. \( k - 3 = 0 \) → \( k = 3 \)
3. \( k + 5 = 0 \) → \( k = -5 \)

### Step 9: Check for concurrency
However, we need to check if \( k = 2 \) leads to parallel lines. Substituting \( k = 2 \) into the first two equations yields:

1. \( 2x + 2y + 2 = 0 \) (or \( x + y + 1 = 0 \))
2. \( 2x + 2y + 3 = 0 \) (or \( x + y + 1.5 = 0 \))

These lines are parallel and thus cannot be concurrent.

### Final Result
The valid values of \( k \) for which the lines are concurrent are:

\[ \boxed{3 \text{ and } -5} \]