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Mathematics
Problem
If the system of linear equations
x
−
2
y
+
k
z
=
1
x - 2y + kz = 1
x
−
2
y
+
k
z
=
1
,
2
x
+
y
+
z
=
2
2x + y + z = 2
2
x
+
y
+
z
=
2
,
3
x
−
y
−
k
z
=
3
3x - y - kz = 3
3
x
−
y
−
k
z
=
3
has a solution
(
x
,
y
,
z
)
(x, y, z)
(
x
,
y
,
z
)
,
z
≠
0
z \neq 0
z
=
0
, then
(
x
,
y
)
(x, y)
(
x
,
y
)
lies on the straight line whose equation is : [JEE-MAIN ONLINE-2019] (1)
3
x
−
4
y
−
1
=
0
3x - 4y - 1 = 0
3
x
−
4
y
−
1
=
0
(2)
3
x
−
4
y
−
4
=
0
3x - 4y - 4 = 0
3
x
−
4
y
−
4
=
0
(3)
4
x
−
3
y
−
4
=
0
4x - 3y - 4 = 0
4
x
−
3
y
−
4
=
0
(4)
4
x
−
3
y
−
1
=
0
4x - 3y - 1 = 0
4
x
−
3
y
−
1
=
0
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