Q.

The projection of line 3x-y+2z-1=0=x+2y-z-2 on the plane 3x+2y+z=0 is

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a

x+111=y−1−9=z−1−15

b

3x−8y+7z+4=0=3x+2y+z

c

x+1211=y+8−9=z+1415

d

x+1211=y+8−9=z+14−15

answer is A.

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Detailed Solution

Equation of a plane passing through the line 3x−y+2z−1=0=x+2y−z−2 is 3x−y+2z−1+λ(x+2y−z−2)=0 Since it is perpendicualr to the given plane ∴λ=−32     a1a2+b1b2+c1c2=0⇒λ+33+2λ-12+-λ+2=0 Equation of the line of projection is 3x−8y+7z+4=0=3x+2y+z Its direction ratios are <11,−9,−15> and the point (−1,1,1) lies on the line x+111=y−1−9=z−1−15 is also the equation of the line of projection.
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