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:
Find the equation of plane passing through the points P(1, 1, 0), Q(1, 2, 1) and R(−2, 2, −1). Choose the correct option.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2x+3y-3z=5\)
Two vectors in the plane:
\(\overrightarrow{PQ} = (0, 1, 1)\)
\(\overrightarrow{PR} = (-3, 1, -1)\)
Their cross product is a normal to the plane:
\( \displaystyle \overrightarrow{PQ}\times\overrightarrow{PR} = \begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\ 0 & 1 & 1\\ -3 & 1 & -1\end{vmatrix} \)
\(= (-1-1)\hat{\imath}-(0+3)\hat{\jmath}+(0+3)\hat{k}\)
\(= -2\hat{\imath}-3\hat{\jmath}+3\hat{k}\)
Taking the normal as (2, 3, −3), the plane through P(1, 1, 0) is
\(2(x-1)+3(y-1)-3(z-0) = 0\)
\(2x+3y-3z = 5\)
Check with R(−2, 2, −1): \(-4+6+3 = 5\).
Note: This topic was not in the CUET 2026 syllabus.