Given equations are:
3x – y – 3 = 0 ... (1)
2x + y – 12 = 0 ... (2)
x – 2y – 1 = 0 ... (3)

To Solve (1) and (2),
(1) + (2) ⇒ 5x = 15 ⇒ x = 3
(2) ⇒ y = 12 - 6 = 6
Thus (1) and (2) intersect at C (3, 6).
To solve (2) and (3),
(2) - 2 (3) ⇒ 5y = 10 ⇒ y = 2
(2) ⇒ 2x = 12 - 2 = 10 ⇒ x = 5
Thus (2) and (3) intersect at B (5, 2).
To solve (3) and (1),
2 (1) - (3) ⇒ 5x = 5 ⇒ x = 1
(3) ⇒ 1 - 2y = 1 ⇒ y = 0
Thus (3) and (1) intersect at A(1, 0).
Area =

3

∫

1

(3x - 3) dx +

5

∫

3

(12 - 2x) dx -

5

∫

1

1

2

(x - 1) dx
= 3 [

x2

2

−x]13 + [12x−x2]35 -

1

2

[

x2

2

−x]15
= 3 [(

9

2

−3)−(

1

2

−1)] + [(60 - 25) - (36 - 9)] -

1

2

[(

25

2

−5)−(

1

2

−1)]
= 3 [

3

2

+

1

2

] + [35 - 27] -

1

2

[

15

2

+

1

2

]
= 6 + 8 - 4 = 10 sq. units