If the normal to the curve y(x)=∫ _0^x(2t^2-15t+10)dt at a point (a,b) is parallel to the line x+3y=-5,a>1, then the value of |a+6b| is equal to .

$$y(x)={\int }_{0}^{x}(2{t}^{2}-15t+10)dt$$y(x)=∫0x​(2t2−15t+10)dt

$${{y}^{'}(x)]}_{x=a}={[2{x}^{2}-15x+10]}_{a}=2{a}^{2}-15a+10$$y′(x)]x=a​=[2x2−15x+10]a​=2a2−15a+10

Slope of normal $=-\frac{1}{3}$=−31​

$$\Rightarrow 2{a}^{2}-15a+10=3\Rightarrow a=7$$⇒2a2−15a+10=3⇒a=7

and $a=\frac{1}{2}$a=21​ (rejected)

$$b=y(7)={\int }_{0}^{7}(2{t}^{2}-15t+10)dt$$b=y(7)=∫07​(2t2−15t+10)dt

$$={[\frac{2{t}^{3}}{3}-\frac{15{t}^{2}}{2}+10t]}_{0}^{7}$$=[32t3​−215t2​+10t]07​

$$\Rightarrow 6b=4\times {7}^{3}-45\times 49+60\times 7$$⇒6b=4×73−45×49+60×7

$$|a+6b|=406$$∣a+6b∣=406