Let A≡(1,3),B≡(5,0),C≡(−1,2) For point A(1,3),3x+2y=3+6=9>0 For point B(5,0),3x+2y=15+0=15>0 For point C(−1,2),3x+2y=−3+4=1>0 ∴ Option (a) is rejected. Now, for option (c) 2x−3y−12=2−9−12<0 ∴ Option (c) is rejected. Also, 2x+y−13=4+3−13<0 ∴ Option (d) is rejected. So, option (b) is correct.