Let A(5,−3)B(3,−2)C(−1,5) be three points. Let P(h,k) satisfying ‌(PA)2+2(PB)2=3(PC)2 ‌(h−5)2+(k+3)2+2(h−3)2+2(k+2)2 ‌=3(h+1)2+3(k−5)2 ‌⇒−10h+25+6k+9−12h+18+8k+8 ‌=6h+3−30k+75 ‌⇒−28h+44k−18=0⇒14h−22k+9=0 ‌‌ Locus ‌14x−22y+9=0 Only option (d), (2,‌