Answer:
The equation of the normal is y = 4 - 2·x
Step-by-step explanation:
The given equation of the circle, is presented as follows;
(x - 3)² + (y + 2)² = 20
The point of the normal of the circle = (1, 2)
The equation of the normal to a circle, x² + y² + D·x + E·y + F = 0 at a point P(x₁, y₁) is given as follows;

Expanding the given equation of the circle, gives;
(x - 3)² + (y + 2)² = x² + y² - 6·x + 4·y + 13 = 20
∴ x² + y² - 6·x + 4·y + 13 - 20 = x² + y² - 6·x + 4·y - 7 = 0
x² + y² - 6·x + 4·y - 7 = 0
∴ x₁ = 1, y₁ = 2, D = -6, E = 4, and F = 13
Which gives;

∴ y - 2 = -2 × (x - 1) = 2 - 2·x
y = 2 - 2·x + 2 = 4 - 2·x
The equation of the normal to the circle with equation (x - 3)² + (y + 2)² = 20, at the point (1, 2) is y = 4 - 2·x