If the normal to the curve y x = ∫ 0 x 2 t 2 - 15 t + 10 d t at a point a , b is parallel to the line…

$y\left(x\right)={\int }_0^x\left(2t^2-15t+10\right)dt$

${\left.y^{\text{'}}\left(x\right)\right]}_{x=a}={\left[2x^2-15x+10\right]}_a=2a^2-15a+10$

Slope of normal $=-\frac{1}{3}$

$\Rightarrow 2a^2-15a+10=3\Rightarrow a=7$

and $a=\frac{1}{2}$  (rejected)

$b=y\left(7\right)={\int }_0^7\left(2t^2-15t+10\right)dt$

$={\left[\frac{2t^3}{3}-\frac{15t^2}{2}+10t\right]}_0^7$

$\Rightarrow 6b=4\times 7^3-45\times 49+60\times 7$

$\left|a+6b\right|=406$