If Q(0,−1,−3) is the image of the point P in the plane 3x−y+4z=2 and R is the point (3,−1,−2), then the area (in sq. units) of ΔPQR is :
A
√914
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B
2√13
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C
√652
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D
√912
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Solution
The correct option is D√912 Let M be a point on the plane such that PQ passes through M.
The point R lies on the plane : 3x−y+4z−2=0 PM=∣∣∣1−12−2√9+1+16∣∣∣=√132 PR=√9+1=√10
Now , using pythagorus theorem in ΔPMR, we have RM=√10−132=√72 ⇒ΔPQR=2×12×√132×√72=√912