A curve C is described by the equation 3x^2 - 2y^2 + 2x - 3y + 5 = 0 - Edexcel - A-Level Maths Pure - Question 3 - 2006 - Paper 6

Next, we substitute the point (0, 1) into the derivative: At (0, 1): $6(0) + 2 - (4(1) + 3) \frac{dy}{dx} = 0$ This simplifies to: $2 - 7 \frac{dy}{dx} = 0$ Thus: $\frac{dy}{dx} = \frac{2}{7}$

The slope of the normal line is the negative reciprocal of the derivative: $m_{normal} = -\frac{1}{\frac{2}{7}} = -\frac{7}{2}$

Using the point-slope form of the equation of a line, we have: $y - 1 = m_{normal}(x - 0)$ Substituting the value of the normal slope: $y - 1 = -\frac{7}{2}(x)$ This can be rearranged to: $7x + 2y - 2 = 0$ Thus, in the form ax + by + c = 0, we have a = 7, b = 2, c = -2.