A team of 5 persons goes on a rescue operation in a helicopter. They know the coordinates of places from where people need to be rescued. These are P(1, 1, 0), Q(1, 2, 1) and R(−2, 2, −1). They decide to stop at Point A (2, 1, 3). From A, 3 team members are dropped on ground using a rope. They reach the point B at the ground level so that distance AB is shortest from the plane passing through P, Q and R.
Based on the above information, answer the following:
The rescue team can rescue people at the same plane (as found previously). If there are more people at the following points, which of these can be rescued by the team?
(A) (1, 3, 2)
(B) (1, 1, 1)
(C) \(\left(\displaystyle\frac{5}{2},1,1\right)\)
(D) \(\left(2,\displaystyle\frac{7}{3},2\right)\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (C) and (D) only
The plane through P(1, 1, 0), Q(1, 2, 1) and R(−2, 2, −1) has normal
\(\overrightarrow{PQ}\times\overrightarrow{PR} = (0, 1, 1)\times(-3, 1, -1) = (-2, -3, 3)\)
so its equation is
\(2x+3y-3z = 5\)
So a point can be rescued exactly when it satisfies 2x + 3y − 3z = 5.
Checking the statements
(A) (1, 3, 2) — Correct. \(2+9-6 = 5\).
(B) (1, 1, 1) — Incorrect. \(2+3-3 = 2 \ne 5\).
(C) \(\left(\frac{5}{2},1,1\right)\) — Correct. \(5+3-3 = 5\).
(D) \(\left(2,\frac{7}{3},2\right)\) — Correct. \(4+7-6 = 5\).
Note: This topic was not in the CUET 2026 syllabus.