The user wants me to solve a matching question from Coordinate Geometry involving three lines: $L_1: x + 3y - 5 = 0$ $L_2: 3x - ky - 1 = 0$ $L_3: 5x + 2y - 12 = 0$

I need to find the value(s) of 'k' for four different conditions described in Column I and match them with the values in Column II.

Column I: (A) $L_1, L_2, L_3$ are concurrent. (B) One of $L_1, L_2, L_3$ is parallel to at least one of the other two. (C) $L_1, L_2, L_3$ form a triangle. (D) $L_1, L_2, L_3$ do not form a triangle.

Column II: (P) $k = -9$ (Q) $k = -6/5$ (R) $k = 5/6$ (S) $k = 5$

Let's analyze each condition in Column I step-by-step.

(A) $L_1, L_2, L_3$ are concurrent

For three lines to be concurrent, they must intersect at a single point. This can be checked in two ways:

Method 1: Using Determinants The condition for three lines $a_1x+b_1y+c_1=0$, $a_2x+b_2y+c_2=0$, and $a_3x+b_3y+c_3=0$ to be concurrent is that the determinant of their coefficients is zero: $\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$ For the given lines, this becomes: $\begin{vmatrix} 1 & 3 & -5 \\ 3 & -k & -1 \\ 5 & 2 & -12 \end{vmatrix} = 0$ Expanding the determinant: $1((-k)(-12) - (2)(-1)) - 3((3)(-12) - (5)(-1)) + (-5)((3)(2) - (5)(-k)) = 0$ $1(12k + 2) - 3(-36 + 5) - 5(6 + 5k) = 0$ $12k + 2 - 3(-31) - 30 - 25k = 0$ $12k + 2 + 93 - 30 - 25k = 0$ $-13k + 65 = 0$ $13k = 65$ $k = 5$

Method 2: Finding intersection point First, find the intersection point of $L_1$ and $L_3$ (since they don't contain $k$). $L_1: x + 3y = 5 \implies x = 5 - 3y$ $L_3: 5x + 2y = 12$ Substitute $x$ from $L_1$ into $L_3$: $5(5 - 3y) + 2y = 12$ $25 - 15y + 2y = 12$ $13 = 13y \implies y = 1$ Now find $x$: $x = 5 - 3(1) = 2$. The intersection point of $L_1$ and $L_3$ is $(2, 1)$. For the lines to be concurrent, this point must also lie on $L_2$. $L_2: 3x - ky - 1 = 0$ Substitute $(x,y) = (2,1)$: $3(2) - k(1) - 1 = 0$ $6 - k - 1 = 0$ $5 - k = 0 \implies k = 5$

Both methods give $k=5$. This corresponds to (S) in Column II. Therefore, (A) maps to (S).

(B) One of $L_1, L_2, L_3$ is parallel to at least one of the other two

Two lines are parallel if their slopes are equal. Let's find the slopes of the three lines. Slope of $L_1 (m_1) = -\frac{\text{coeff of } x}{\text{coeff of } y} = -\frac{1}{3}$ Slope of $L_2 (m_2) = -\frac{3}{-k} = \frac{3}{k}$ Slope of $L_3 (m_3) = -\frac{5}{2}$

Now we check for parallelism between pairs of lines:

1. $L_1 \|\| L_2$: $m_1 = m_2$ $-\frac{1}{3} = \frac{3}{k} \implies -k = 9 \implies k = -9$. This corresponds to (P).
2. $L_2 \|\| L_3$: $m_2 = m_3$ $\frac{3}{k} = -\frac{5}{2} \implies 6 = -5k \implies k = -\frac{6}{5}$. This corresponds to (Q).
3. $L_1 \|\| L_3$: $m_1 = m_3$ $-\frac{1}{3} = -\frac{5}{2}$, which is false. So $L_1$ and $L_3$ are never parallel.

So, one line is parallel to another if $k = -9$ or $k = -6/5$. Therefore, (B) maps to (P) and (Q).

(D) $L_1, L_2, L_3$ do not form a triangle

Three lines in a plane do not form a triangle if either:

1. At least two of the lines are parallel.
2. The three lines are concurrent (intersect at a single point).

From our analysis in (A) and (B):

• The lines are concurrent if $k = 5$.
• At least two lines are parallel if $k = -9$ or $k = -6/5$.

Combining these conditions, the lines do not form a triangle if $k \in \{-9, -6/5, 5\}$. These values correspond to (P), (Q), and (S) in Column II. Therefore, (D) maps to (P), (Q), and (S).

(C) $L_1, L_2, L_3$ form a triangle

Three lines form a triangle if and only if they are not concurrent and no two lines are parallel. This is the exact opposite of condition (D). So, a triangle is formed if $k \notin \{-9, -6/5, 5\}$.

We need to check which value(s) from Column II satisfy this condition.

• (P) $k = -9$: Does not form a triangle.
• (Q) $k = -6/5$: Does not form a triangle.
• (R) $k = 5/6$: This value is not in the set $\{-9, -6/5, 5\}$. So, for this value, the lines form a triangle.
• (S) $k = 5$: Does not form a triangle.

Thus, among the given options in Column II, only $k=5/6$ results in the lines forming a triangle. Therefore, (C) maps to (R).

Summary of Matches

• (A) $\to$ (S)
• (B) $\to$ (P), (Q)
• (C) $\to$ (R)
• (D) $\to$ (P), (Q), (S)

Let's assume the mapping P=i, Q=ii, R=iii, S=iv for the multiple choice options:

• A $\to$ iv
• B $\to$ i, ii
• C $\to$ iii
• D $\to$ i, ii, iv

Now, we compare this with the given options: A: A - iv; B - ii; C - iii; D - i, ii (Incorrect) B: A - iv; B - i, ii; C - iii; D - i, ii, iv (Correct) C: A - iv; B - i; C - iii; D - i, ii (Incorrect) D: A - ii; B - i, iii; C - iii; D - i, ii, iv (Incorrect)

The correct choice is B.