If the normal to the curve $$y(x) = \int_0^x (2t^2 - 15t + 10)\,dt$$ at a point $$(a, b)$$ is parallel to the line $$x + 3y = -5, a > 1$$, then the value of $$|a + 6b|$$ is equal to ________.

We have $$y(x) = \int_0^x (2t^2 - 15t + 10)\,dt$$. By the Fundamental Theorem of Calculus, $$y'(x) = 2x^2 - 15x + 10$$.

The normal to the curve at $$(a, b)$$ is parallel to $$x + 3y = -5$$, whose slope is $$-\frac{1}{3}$$. The slope of the normal is $$-\frac{1}{y'(a)}$$, so $$-\frac{1}{y'(a)} = -\frac{1}{3}$$, giving $$y'(a) = 3$$.

Setting $$2a^2 - 15a + 10 = 3$$: $$2a^2 - 15a + 7 = 0$$. Using the quadratic formula: $$a = \frac{15 \pm \sqrt{225 - 56}}{4} = \frac{15 \pm \sqrt{169}}{4} = \frac{15 \pm 13}{4}$$.

So $$a = 7$$ or $$a = \frac{1}{2}$$. Since $$a > 1$$, we have $$a = 7$$.

Now $$b = y(7) = \int_0^7 (2t^2 - 15t + 10)\,dt = \left[\frac{2t^3}{3} - \frac{15t^2}{2} + 10t\right]_0^7$$.

Evaluating: $$\frac{2(343)}{3} - \frac{15(49)}{2} + 70 = \frac{686}{3} - \frac{735}{2} + 70$$.

Finding a common denominator of 6: $$\frac{1372}{6} - \frac{2205}{6} + \frac{420}{6} = \frac{1372 - 2205 + 420}{6} = \frac{-413}{6}$$.

Therefore $$|a + 6b| = \left|7 + 6 \cdot \frac{-413}{6}\right| = |7 - 413| = |-406| = 406$$.