We have the lines: L1 : x + 3y - 5 = 0 L2 : 3x - ky - 1 = 0 L3 : 5x + 2y - 12 = 0 (A)→(S)The point of the intersection of L1 and L3 is (2, 1). If the lines are concurrent, then (2, 1) satisfy L2. Therefore, 3(2) - (1) -1 = 0⇒ = 5 (B) → (P), (Q) If L1 and L2 are parallel:
3
1
=
−k
3
≠
−1
−5
⇒ k = - 9 If L2 and L3 are parallel: Hence,
3
5
= −
k
2
⇒ k = −
6
5
(C)→(R) Since the lines form a triangle, they cannot be concurrent and no two of them are parallel ⇒ k ≠ 5 , - 9 , −
6
5
⇒ k =
5
6
(D)→(P), (Q), (S) If the lines do not form a triangle, k = 5 , −9,−