We are given the equation of the curve:

$y = 2x^3 - 3x^2 + 3$

Additionally, the line $y = 12x + 5$ is parallel to the tangents at two points, $P$ and $Q$, on the curve.

Step 1: Finding the slope of the tangent line

Since the tangents are parallel to the line $y = 12x + 5$, the slope of the tangents at $P$ and $Q$ must equal the slope of this line. The slope of the line $y = 12x + 5$ is 12, as it is the coefficient of $x$.

Step 2: Derivative of the curve to find the slope of the tangents

The slope of the tangent to the curve at any point is given by the derivative of the curve equation. Let's compute the derivative of $y = 2x^3 - 3x^2 + 3$:

$\frac{dy}{dx} = 6x^2 - 6x$

Step 3: Set the derivative equal to the slope of the line

We set the derivative equal to the slope of the tangent, which is 12:

$6x^2 - 6x = 12$

Simplifying the equation:

$x^2 - x - 2 = 0$

Factoring this quadratic equation:

$(x - 2)(x + 1) = 0$

So, $x = 2$ or $x = -1$.

Step 4: Find the coordinates of $P$ and $Q$

To find the coordinates of $P$ and $Q$, substitute these values of $x$ back into the original curve equation $y = 2x^3 - 3x^2 + 3$.

For $x = 2$:

$y = 2(2)^3 - 3(2)^2 + 3 = 16 - 12 + 3 = 7$

So, the coordinates of $Q$ are $(2, 7)$.

For $x = -1$:

$y = 2(-1)^3 - 3(-1)^2 + 3 = -2 - 3 + 3 = -2$

So, the coordinates of $P$ are $(-1, -2)$.

Thus, the coordinates of $P$ and $Q$ are $(-1, -2)$ and $(2, 7)$, respectively.

Step 5: Find the equation of the tangent at $Q$

We already know that the slope of the tangent at $Q$ is 12 (from the parallel condition). The equation of a tangent line is given by the point-slope form:

$y - y_1 = m(x - x_1)$

Using the point $Q(2, 7)$ and slope $m = 12$:

$y - 7 = 12(x - 2)$

Expanding this:

$y - 7 = 12x - 24$ $y = 12x - 17$

Thus, the equation of the tangent at $Q$ is:

$y = 12x - 17$

Would you like any further details or clarification on this solution?

Here are some related questions for further understanding:

1. How would the solution change if the slope of the line was different?
2. What is the geometric interpretation of the points $P$ and $Q$ on the curve?
3. How do you find the equation of a tangent line at a general point on a curve?
4. Can you generalize this approach to find tangents parallel to any given line?
5. What happens if the curve has more than two points with the same slope?

Tip: Always compute the derivative of the curve first when dealing with tangent lines. This helps in finding the slope at any point on the curve.