Given that we need to find the equation of the circle formed by the lines:

⇒ x + y = 2

⇒ 3x - 4y = 6

⇒ x - y = 0

On solving these lines we get the intersection points A(2,0), B(- 6, - 6), C(1,1)

We know that the standard form of the equation of a circle is given by:

⇒ x² + y² + 2ax + 2by + c = 0 .....(1)

Substituting (2,0) in (1), we get

⇒ 2² + 0² + 2a(2) + 2b(0) + c = 0

⇒ 4 + 4a + c = 0

⇒ 4a + c + 4 = 0 ..... (2)

Substituting (- 6, - 6) in (1), we get

⇒ (- 6)² + (- 6)² + 2a(- 6) + 2b(- 6) + c = 0

⇒ 36 + 36 - 12a - 12b + c = 0

⇒ 12a + 12b - c - 72 = 0 ..... (3)

Substituting (1,1) in (1), we get

⇒ 1² + 1² + 2a(1) + 2b(1) + c = 0

⇒ 1 + 1 + 2a + 2b + c = 0

⇒ 2a + 2b + c + 2 = 0 ..... (4)

Solving (2), (3), (4) we get

⇒ a = 2, b = 3,c = - 12.

Substituting these values in (1), we get

⇒ x² + y² + 2(2)x + 2(3)y - 12 = 0

⇒ x² + y² + 4x + 6y - 12 = 0

∴ The equation of the circle is x² + y² + 4x + 6y - 12 = 0.