If the line x−23=y+12=z−1−1 intersects the plane 2x+3y-z+13=0 at a point P then PQ is equal to

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answer is D.

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Detailed Solution

x−23=y+12=z−1−1=tP(3t1+2,2t1−1,−t1+1)1 lies on 2x+3y-z+13=0t1=−1⇒P(−1,−3,2)Q(3t2+2,2t2−1,−t2+1) lies on 3x+y+4z=16⇒t2=1⇒Q(5,1,0)

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