    Question

# The equation of the plane containing the lines 2x−5y+z=3, x+y+4z=5 and parallel to the plane x+3y+6z=1, is:

A
2x+6y+12z=13
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B
x+3y+6z=7
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C
x+3y+6z=7
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D
2x+6y+12z=13
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Solution

## The correct option is C x+3y+6z=7Given : p1:2x−5y+z−3=0 and p2:x+y+4z−5=0 Any plane through the line of intersection is p1+λp2=0 The equation of required plane can be obtained as: 2x−5y+z−3+λ(x+y+4z−5)=0⋯(i) As plane is parallel to x+3y+6z−1=0, 2+λ1=λ−53=1+4λ6 ⇒6+3λ=λ−5⇒λ=−112 Substituting the value of λ in (i), we have −7x−21y−42z+49=0 ⇒x+3y+6z−7=0  Suggest Corrections  0      Similar questions
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