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Q.

If the system of linear equations x−2y+kz=1     2x+y+z=2,   3x−y−kz=3has a solution (x,y,z),  z≠0,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

answer is C.

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Detailed Solution

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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