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If the system of linear equations x2y+kz=1     2x+y+z=2,   3xykz=3has a solution (x,y,z),  z0,Then (x,y)lies on the straight line whose equation is

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a
3x−4y−1=0
b
3x−4y−4=0
c
4x−3y−4=0
d
4x−3y−1=0

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detailed solution

Correct option is C

x−2y+kz=1→(1)2x+y+z=2→(2)3x−y−kz=3→(3)(1)+(3)⇒4x−3y=4⇒(x, y)lies on the straight line 4x−3y−4=0


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