Using the method of integration, find the area of the region bounded by the following lines: 3x-y=0, 2x+y=0, x-y=0

Given lines are
3x-y=0—(1)
2x+y=0—(2)
x-y=0—(3)
For intersecting point of (1) and (2)
(1) + (2) ⇒3x-y+2x+y=0⇒5x=0⇒x=0
Putting x = 0 in (1), we get
0-y=0
y=0
Intersecting point of (1) and (2) is (0, 0)
For intersecting point of (2) and (3)
(2)-(3) ⇒2x+y-(x-y)=0⇒x+2y=0
Putting y=0 in (2), we get 2x=0, x=0
Intersecting point of (2) and (3) is (0, 0)
For intersecting point of (1) and (3)
(1)-(3) ⇒3x-y-(x-y)=0⇒2x=0⇒x=0
Putting x=0 in (1), we get -y=0, y=0
Intersecting point (1) and (3) is (0, 0)
There must be some error in the question.
Let us assume the lines are
3x-y=0
2x+y=0
x=0
For intersecting point of (1) and (2)
(1)+(2) ⇒3x-y+2x+y=0⇒5x=0⇒x=0
Putting x=0 in (1), we get y=0
Intersecting point of (1) and (2) is (0, 0)
For intersecting point of (2) and (3)
Putting x=0 in (2), we get y=0
Intersecting point of (2) and (3) is (0, 0)
For intersecting point of (1) and (3)
Putting x=0 in (1), we get y=0
Intersecting point of (1) and (3) is (0, 0)
Let's assume another set of lines:
3x-y=3
2x+y=0
x=0
For intersecting point of (1) and (2)
(1)+(2) ⇒5x=3⇒x=3/5
Putting x=3/5 in (2), we get y=-6/5
Intersecting point of (1) and (2) is (3/5, -6/5)
For intersecting point of (2) and (3)
Putting x=0 in (2), we get y=0
Intersecting point of (2) and (3) is (0, 0)
For intersecting point of (1) and (3)
Putting x=0 in (1), we get y=-3
Intersecting point of (1) and (3) is (0, -3)