The length of the perpendicular from the origin, on normal to the curve, $$x^2 + 2xy - 3y^2 = 0$$, at the point (2, 2), is.

We have the implicit curve $$x^2 + 2xy - 3y^2 = 0$$ and the specific point $$(2,2)$$ on it. To obtain the slope of the tangent at this point, we first differentiate the given equation with respect to $$x$$, remembering that $$y = y(x)$$ is a function of $$x$$.

Differentiating term by term:

$$\dfrac{d}{dx}(x^2) + \dfrac{d}{dx}(2xy) - \dfrac{d}{dx}(3y^2) = 0.$$

Using $$\dfrac{d}{dx}(x^2) = 2x$$, the product rule $$\dfrac{d}{dx}(2xy) = 2y + 2x\dfrac{dy}{dx}$$, and the chain rule $$\dfrac{d}{dx}(3y^2) = 6y\dfrac{dy}{dx}$$, we get

$$2x + (2y + 2x\dfrac{dy}{dx}) - 6y\dfrac{dy}{dx} = 0.$$

Collecting the $$\dfrac{dy}{dx}$$ terms together,

$$2x + 2y + (2x - 6y)\dfrac{dy}{dx} = 0.$$

Solving for the derivative $$\dfrac{dy}{dx}$$ (which is the slope of the tangent), we have

$$(2x - 6y)\dfrac{dy}{dx} = -\,(2x + 2y),$$

so

$$\dfrac{dy}{dx} = -\dfrac{2x + 2y}{2x - 6y}.$$

Now substitute the coordinates $$(2,2)$$:

$$\dfrac{dy}{dx}\Big|_{(2,2)} = -\dfrac{2\cdot 2 + 2\cdot 2}{2\cdot 2 - 6\cdot 2} = -\dfrac{4 + 4}{4 - 12} = -\dfrac{8}{-8} = 1.$$

Thus the slope of the tangent at $$(2,2)$$ is $$m_t = 1$$. The slope of the normal is the negative reciprocal, therefore

$$m_n = -\dfrac{1}{m_t} = -1.$$

The normal line passing through $$(2,2)$$ with slope $$-1$$ is obtained from the point-slope form $$y - y_1 = m(x - x_1)$$:

$$y - 2 = -1(x - 2).$$

Simplifying,

$$y - 2 = -x + 2 \quad\Longrightarrow\quad x + y - 4 = 0.$$

We must now find the perpendicular distance of the origin $$(0,0)$$ from this normal line. For a line written as $$Ax + By + C = 0$$, the distance from a point $$(x_0,y_0)$$ is given by the formula

$$\text{Distance} = \dfrac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}.$$

Here $$A = 1,\; B = 1,\; C = -4$$ and $$(x_0,y_0) = (0,0).$$ Substituting, we get

$$\text{Distance} = \dfrac{|1\cdot 0 + 1\cdot 0 - 4|}{\sqrt{1^2 + 1^2}} = \dfrac{|-4|}{\sqrt{2}} = \dfrac{4}{\sqrt{2}} = 4 \times \dfrac{\sqrt{2}}{2} = 2\sqrt{2}.$$

Hence, the correct answer is Option D.