The curve C has equation $y = \frac{1}{3}x^3 - 4x^2 + 8x + 3$ - Edexcel - A-Level Maths Pure - Question 1 - 2005 - Paper 1

First, we need to find the derivative of the curve to obtain the slope of the tangent:

$\frac{dy}{dx} = \frac{1}{3}(3x^2) - 8x + 8 = x^2 - 8x + 8.$

Now, we calculate the slope at $x = 3$:

$\frac{dy}{dx} \bigg|_{x=3} = (3)^2 - 8(3) + 8 = 9 - 24 + 8 = -7.$

The slope $m = -7$.

Using the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$, with the point (3, 0):

$y - 0 = -7(x - 3)$

Thus the equation of the tangent line is:

$y = -7x + 21.$

To find the coordinates of point Q where the tangent is parallel to the tangent at P, we require the slope to be equal to -7.

We set our earlier derivative equal to -7:

$x^2 - 8x + 8 = -7$

This simplifies to:

$x^2 - 8x + 15 = 0.$

Now we can factor this quadratic:

$(x - 3)(x - 5) = 0$

Thus, $x = 3$ or $x = 5$. Since we are looking for another point Q, we choose $x = 5$.

Now we substitute $x = 5$ back into the original curve equation to find $y$:

$y = \frac{1}{3}(5)^3 - 4(5)^2 + 8(5) + 3$

Calculating this gives:

$y = \frac{1}{3}(125) - 100 + 40 + 3 = \frac{125}{3} - 100 + 40 + 3.$

Converting -100 to common denominator:

$y = \frac{125}{3} - \frac{300}{3} + \frac{120}{3} + \frac{9}{3} = \frac{125 - 300 + 120 + 9}{3} = \frac{-46}{3}.$

Therefore, the coordinates of point Q are $Q(5, \frac{-46}{3})$.