1. **Problem statement:** Find the point on the curve $y = \sqrt{2x + 1}$ where the normal line is parallel to the line $y = -3x + 6$. 2. **Understanding the problem:** The slope of the given line is $-3$. Since the normal to the curve is parallel to this line, the slope of the normal to the curve must also be $-3$. 3. **Find the slope of the tangent to the curve:** Given $y = \sqrt{2x + 1} = (2x + 1)^{1/2}$. Using the chain rule: $$\frac{dy}{dx} = \frac{1}{2}(2x + 1)^{-1/2} \times 2 = \frac{1}{\sqrt{2x + 1}}$$ 4. **Slope of the normal:** The normal is perpendicular to the tangent, so its slope $m_n$ satisfies: $$m_n = -\frac{1}{m_t}$$ where $m_t$ is the slope of the tangent. 5. **Set the slope of the normal equal to $-3$:** $$-\frac{1}{m_t} = -3 \implies m_t = \frac{1}{3}$$ 6. **Find $x$ such that $m_t = \frac{1}{3}$:** $$\frac{1}{\sqrt{2x + 1}} = \frac{1}{3} \implies \sqrt{2x + 1} = 3$$ Square both sides: $$2x + 1 = 9 \implies 2x = 8 \implies x = 4$$ 7. **Find corresponding $y$ value:** $$y = \sqrt{2(4) + 1} = \sqrt{8 + 1} = \sqrt{9} = 3$$ 8. **Answer:** The point is $(4, 3)$. This corresponds to option (A).