Given that we need to find the equation of the circle with centre (1, - 2) and passing through the point of intersection of the lines 3x + y = 14 and 2x + 5y = 18.

Let us first find the point of intersection of lines 3x + y = 14 and 2x + 5y = 18.

On solving the lines, we get the intersection point to be (4,2).

We have a circle with centre (1, - 2) and passing through the point (4,2).

We know that the radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.

We know that the distance between the two points (x₁,y₁) and (x₂,y₂) is .

Let us assume r is the radius of the circle.

⇒

⇒

⇒

⇒ r = √25

⇒ r = = 5 ..... (1)

We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by:

⇒ (x - p)² + (y - q)² = r²

Now we substitute the corresponding values in the equation:

⇒ (x - 1)² + (y - ( - 2))² = (5)²

⇒ (x - 1)² + (y + 2)² = 25

⇒ x² - 2x + 1 + y² + 4y + 4 = 25

⇒ x² + y² - 2x + 4y - 20 = 0.

∴The equation of the circle is x² + y² - 2x + 4y - 20 = 0.