Find an equation of the normal line to the parabola y=x^2-5x+4 that is parallel to the line x-3y=5

The equation of the normal line is parallel to x - 3y = 5
3y = x - 5
y = 1/3x - 5/3
The normal line has a slope of 1/3 and the slope of the tangent line is -3
Let the point at which the normal line passes the parabola be (x1, y1), then
f'(x1) = 2x - 5 = -3
2x = -3 + 5 = 2
x = 2/2 = 1
y = (1)^2 - 5(1) + 4 = 1 - 5 + 4 = 0
Therefore, the normal line passes through point (1, 0) and has a slope of 1/3.
Therefore, required equation is y = 1/3(x - 1)
y = 1/3x - 1/3

It’s d for sure i just did also that’s why