The equation of the normal to the curve y=x^4-6 x^3+13 x^2-10 x+5 at…

$y=x^4-6 x^3+13 x^2-10 x+5$ 

at  x = 1

$y=1-6+13-10+5=3$

point of intersection (1, 3)

$\frac{dy}{dx}=4x^3-18x^2+26x-10$

so $\left.\frac{dy}{dx}\right]_{x=1}=4-18+26-10=2$

so slope of normal = $\frac{-1}{2}$

equation → $(y-3)=\frac{-1}{2}(x-1)$

so $2y+x-7=0$