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 = $\underset{1}{\overset{3}{\int }}$ (3x - 3) dx + $\underset{3}{\overset{5}{\int }}$ (12 - 2x) dx - $\underset{1}{\overset{5}{\int }}\frac{1}{2}$ (x - 1) dx
= 3 ${\left[\frac{{\mathit{x}}^2}{2}-\mathit{x}\right]}_1^3$ + ${\left[12\mathit{x}-{\mathit{x}}^2\right]}_3^5$ - $\frac{1}{2}{\left[\frac{{\mathit{x}}^2}{2}-\mathit{x}\right]}_1^5$
= 3 $\left[\left(\frac{9}{2}-3\right)-\left(\frac{1}{2}-1\right)\right]$ + [(60 - 25) - (36 - 9)] - $\frac{1}{2}\left[\left(\frac{25}{2}-5\right)-\left(\frac{1}{2}-1\right)\right]$
= 3 $\left[\frac{3}{2}+\frac{1}{2}\right]$ + [35 - 27] - $\frac{1}{2}\left[\frac{15}{2}+\frac{1}{2}\right]$
= 6 + 8 - 4 = 10 sq. units