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Triangle formed with the foot or image of a perpendicular on a plane: its area

This type of question, asked 3 times: 29 Jul 2022, 6 Apr 2023, 10 Apr 2023. All 3 questions below, oldest first.

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  1. Question 1

    29 Jul 2022

    164
    Let QQ be the foot of perpendicular drawn from the point P(1,2,3)P(1,2,3) to the plane x+2y+z=14x + 2y + z = 14. If RR is a point on the plane such that ∠PRQ=60∘\angle PRQ =60^{\circ}, then the area of △PQR\bigtriangleup PQR is equal to:

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  2. Question 2

    6 Apr 2023

    89
    Let the image of the point P(1,2,3)P(1,2,3) in the plane 2x2x −y+z=9- y + z = 9 be QQ. If the coordinates of the point RR are (6,10,7)(6,10,7), then the square of the area of the triangle PQRPQR is

  3. Question 3

    10 Apr 2023

    173
    Let the foot of perpendicular from the point A(4,3A(4,3, 1) on the plane P:x−y+2z+3=0P:x - y + 2z + 3 = 0 be NN. If B(5B(5, α,β),α,β∈Z\alpha,\beta),\alpha,\beta \in Z is a point on plane PP such that the area of the triangle ABNABN in 323\sqrt{2}, then α2+β2+αβ\alpha^{2}+\beta^{2}+ \alpha\beta is equal to