The variable plane (2λ+1)x+(3−λ)y+z=4 always passes through the line
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a
x0=y0=z−41
b
x1=y2=z−4−3
c
x1=y1=z−4−7
d
x1=y2=z−4−7
answer is D.
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Detailed Solution
The plane is x+3y+z−4+λ(2x−y)=0This always passes through the line of intersection of the planes x+3y+z−4=0 and 2x−y=0 Now, 2x−y=0⇒ x1=y2∴ x+3y+z−4=0⇒ x+3⋅2x+z−4=0⇒ 7x+z−4=0⇒ 7x=−(z−4)⇒ x1=z−4−7∴ line is x1=y2=z−4−7