1. **State the problem:** Find the equation of the normal to the curve $y = x^3 - 3x^2 + 2x$ at the point $P(3,6)$. 2. **Find the derivative:** The slope of the tangent to the curve at any $x$ is given by the derivative $$y' = \frac{dy}{dx} = 3x^2 - 6x + 2.$$ 3. **Evaluate the derivative at $x=3$ to find the slope of the tangent:** $$y'(3) = 3(3)^2 - 6(3) + 2 = 27 - 18 + 2 = 11.$$ 4. **Find the slope of the normal:** The normal is perpendicular to the tangent, so its slope is the negative reciprocal of the tangent slope: $$m_{normal} = -\frac{1}{11}.$$ 5. **Write the equation of the normal using point-slope form:** Using the point $P(3,6)$ and slope $m_{normal} = -\frac{1}{11}$, $$y - 6 = -\frac{1}{11}(x - 3).$$ 6. **Simplify the equation:** $$y - 6 = -\frac{1}{11}x + \frac{3}{11}$$ $$y = -\frac{1}{11}x + \frac{3}{11} + 6 = -\frac{1}{11}x + \frac{3}{11} + \frac{66}{11} = -\frac{1}{11}x + \frac{69}{11}.$$ **Final answer:** $$\boxed{y = -\frac{1}{11}x + \frac{69}{11}}.$$