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Question

Let P be the image of the point (3, 1, 7) with respect to the plane x - y + z = 3. Then, the equation of the plane passing through P and containing the straight line x1=y2=z1 is


A

x + y - 3z = 0

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B

3x + z = 0

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C

x - 4y + 7z = 0

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D

2x - y = 0

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Solution

The correct option is C

x - 4y + 7z = 0


Let image of Q(3,1,7) w.r.t. x - y + z = 3 be P(α,β,γ). α31=β11=γ71=2(31+7)12+(1)2+(1)2α3=1β=γ7=4 α=1,β=5,γ=3

Hence, the image of Q(3, 1, 7) is P(-1, 5, 3).
To find equation of plane passing through P(-1, 5, 3) and containing x1=y2=z1

∣ ∣x0 y0 z010 20 1010 50 30∣ ∣=0x(65)y(3+1)+z(5+2)=0x4y+7z=0


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