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If the system of linear equations

$x-2y+kz=1$

$2x+y+z=2$

$3x-y-kz=3$

has a solution $(z,y,z), z\neq 0$ then $(x,y)$ lies on the straight line whose equation is :

Option 1)
$3x-4y-1=0$

Option 2)
$4x-3y-4=0$

Option 3)
$4x-3y-1=0$

Option 4)
$3x-4y-4=0$

Answers (1)

best_answer

$x-2y+kz=1............(1)$

$2x+y+z=2............(2)$

$3x-y-kz=3............(3)$

add $(1)$ and $(3)$

$4x-3y=4$

Option 1)

$3x-4y-1=0$

Option 2)

$4x-3y-4=0$

Option 3)

$4x-3y-1=0$

Option 4)

$3x-4y-4=0$

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