Let the foot of perpendicular of the point P(3,−2,−9) on the plane passing through the points (−1,−2,−3),(9,3,4),(9,−2,1) be Q(α,β,γ). Then the distance of Q from the origin is:
(A)29
(B)35
(C)42✓ Correct
(D)38
Answer: (C)
Solution
A normal to the plane through the three points is (10,5,7)×(10,0,4)=(20,30,−50)∥(2,3,−5), so the plane is 2(x+1)+3(y+2)−5(z+3)=0, i.e. 2x+3y−5z−7=0. The foot of the perpendicular from P(3,−2,−9) satisfies 2x−3=3y+2=−5z+9=−22+32+522(3)+3(−2)−5(−9)−7=−3838=−1. This gives Q(1,−5,−4), so the distance of Q from the origin is 1+25+16=42.
JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Single correct option