The number of points on the curve $$y = 54x^5 - 135x^4 - 70x^3 + 180x^2 + 210x$$ at which the normal lines are parallel to $$x + 90y + 2 = 0$$ is:

Step 1: Find the slope of the given line

The equation of the given line is:

$$x + 90y + 2 = 0$$

Rewriting it in slope-intercept form ($$y = mx + c$$):

$$90y = -x - 2 \implies y = -\frac{1}{90}x - \frac{2}{90}$$

Thus, the slope of this line is $$m = -\frac{1}{90}$$.

Step 2: Formulate the condition for parallel normal lines

Since the normal lines to the curve are parallel to the given line, their slope must be equal to the slope of the line:

$$m_{\text{normal}} = -\frac{1}{90}$$

The relationship between the slope of the tangent ($$\frac{dy}{dx}$$) and the slope of the normal is given by:

$$m_{\text{normal}} = -\frac{1}{\frac{dy}{dx}} = -\frac{1}{90} \implies \frac{dy}{dx} = 90$$

Step 3: Differentiate the given curve equation

The equation of the curve is:

$$y = 54x^5 - 135x^4 - 70x^3 + 180x^2 + 210x$$

Differentiating $$y$$ with respect to $$x$$ gives:

$$\frac{dy}{dx} = 54(5x^4) - 135(4x^3) - 70(3x^2) + 180(2x) + 210$$

$$\frac{dy}{dx} = 270x^4 - 540x^3 - 210x^2 + 360x + 210$$

Step 4: Set the derivative equal to 90 and solve the polynomial equation

Substituting $$\frac{dy}{dx} = 90$$ into the derivative equation:

$$270x^4 - 540x^3 - 210x^2 + 360x + 210 = 90$$

Subtracting 90 from both sides:

$$270x^4 - 540x^3 - 210x^2 + 360x + 120 = 0$$

Dividing the entire equation by 30 to simplify:

$$9x^4 - 18x^3 - 7x^2 + 12x + 4 = 0$$

By inspection, we test integer roots:

- For $$x = 1$$: $$9(1)^4 - 18(1)^3 - 7(1)^2 + 12(1) + 4 = 9 - 18 - 7 + 12 + 4 = 0$$ (So, $$x = 1$$ is a root)

- For $$x = 2$$: $$9(2)^4 - 18(2)^3 - 7(2)^2 + 12(2) + 4 = 144 - 144 - 28 + 24 + 4 = 0$$ (So, $$x = 2$$ is a root)

Since $$x = 1$$ and $$x = 2$$ are roots, $$(x-1)(x-2) = x^2 - 3x + 2$$ is a factor. Dividing the polynomial by this factor yields the remaining quadratic equation:

$$9x^2 + 9x + 2 = 0$$

Factoring the quadratic equation:

$$(3x + 1)(3x + 2) = 0 \implies x = -\frac{1}{3} \quad \text{or} \quad x = -\frac{2}{3}$$

The four distinct real roots for the x-coordinates are $$x = 1, 2, -\frac{1}{3}, -\frac{2}{3}$$. Since all four values are real and distinct, they correspond to four unique points on the curve.

Conclusion:

The number of points on the curve at which the normal lines are parallel to the given line is 4.