Consider the lines given by
`L_(1):x+3y-5=0," "L_(2):3x-ky-1=0" "L_(3):5x+2y-12=0`
`{:(,"Column-I",,"Column-II"),((A),","L_(1)","L_(2)","L_(3)" are concurrent, if",(p),k=-9),((B)," One of "L_(1)","L_(2)","L_(3)" is parallel to at least one of the other two, if",(q),k=-6/5),((C),L_(1)","L_(2)","L_(3)" from a triangle, if",(r),k=5/6),((D),L_(1)","L_(2)","L_(3)" do not from a triangle, if",(s),k=5):}`

Answer

To solve the problem, we need to analyze the three lines given by their equations and determine the conditions under which they are concurrent, parallel, or form a triangle.

### Step 1: Identify the equations of the lines
The equations of the lines are:
1. \( L_1: x + 3y - 5 = 0 \)
2. \( L_2: 3x - ky - 1 = 0 \)
3. \( L_3: 5x + 2y - 12 = 0 \)

### Step 2: Check for concurrency
For the lines to be concurrent, the determinant of the coefficients of \(x\), \(y\), and the constant term must be zero. The determinant can be set up as follows:

\[ \begin{vmatrix} 1 & 3 & -5 \\ 3 & -k & -1 \\ 5 & 2 & -12 \end{vmatrix} = 0 \]

Calculating the determinant:

\[ = 1 \begin{vmatrix} -k & -1 \\ 2 & -12 \end{vmatrix} - 3 \begin{vmatrix} 3 & -1 \\ 5 & -12 \end{vmatrix} + (-5) \begin{vmatrix} 3 & -k \\ 5 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} -k & -1 \\ 2 & -12 \end{vmatrix} = (-k)(-12) - (-1)(2) = 12k + 2 \)

2. \( \begin{vmatrix} 3 & -1 \\ 5 & -12 \end{vmatrix} = (3)(-12) - (-1)(5) = -36 + 5 = -31 \)

3. \( \begin{vmatrix} 3 & -k \\ 5 & 2 \end{vmatrix} = (3)(2) - (-k)(5) = 6 + 5k \)

Substituting back into the determinant:

\[ 1(12k + 2) - 3(-31) - 5(6 + 5k) = 0 \]

Simplifying:

\[ 12k + 2 + 93 - 30 - 25k = 0 \]
\[ -13k + 65 = 0 \]
\[ k = 5 \]

### Step 3: Check for parallel lines
For the lines to be parallel, the ratios of the coefficients of \(x\) and \(y\) must be equal for at least one pair of lines.

1. Comparing \(L_1\) and \(L_2\):
\[ \frac{1}{3} = \frac{3}{-k} \implies k = -9 \]

2. Comparing \(L_2\) and \(L_3\):
\[ \frac{3}{5} = \frac{-k}{2} \implies -k = \frac{6}{5} \implies k = -\frac{6}{5} \]

### Step 4: Check for triangle formation
The lines will form a triangle if they are neither concurrent nor parallel. This means that \(k\) should not be equal to 5, -9, or -6/5.

### Summary of Results
- (A) Lines are concurrent if \(k = 5\).
- (B) At least one line is parallel to another if \(k = -9\) or \(k = -\frac{6}{5}\).
- (C) Lines form a triangle if \(k\) is any value except \(5\), \(-9\), or \(-\frac{6}{5}\).
- (D) Lines do not form a triangle if \(k = 5\), \(-9\), or \(-\frac{6}{5}\).

### Final Answers
- (A) \(k = 5\)
- (B) \(k = -9\) or \(k = -\frac{6}{5}\)
- (C) \(k \neq 5, -9, -\frac{6}{5}\)
- (D) \(k = 5\), \(-9\), or \(-\frac{6}{5}\)