Let O be the vertex of the parabola $$x^{2}=4y$$ and Q be any point on it. Let the locus of the point P, which divides the line segment OQ internally in the ratio 2: 3 be the conic C. Then the equation of the chord of C, which is bisected at the point (1, 2), is:

First, parametrize the given parabola
$$x^2=4y$$

using the standard parameter (t):
Q = (2t,, t^2)

Point (P) divides (OQ) internally in the ratio (2:3). That means
P = $$\frac{2}{5}Q=\left(\frac{4t}{5},,\frac{2t^2}{5}\right)$$

Eliminate (t):
x = $$\frac{4t}{5}\Rightarrow t=\frac{5x}{4}$$
y = $$\frac{2}{5}t^2=\frac{2}{5}\cdot\frac{25x^2}{16}=\frac{5x^2}{8}$$

So the locus (C) is:
$$5x^2=8y$$

Now, we need the equation of the chord of this parabola whose midpoint is ((1,2)).

For a parabola, the chord with midpoint (($$x_1,y_1$$)) is given by the midpoint formula (T = $$S_1$$).

Using the quadratic form:

$$5x^2-8y=0$$

The chord with midpoint ((1,2)) becomes:
$$5x(1)-4(y+2)=5(1)^2-8(2)$$

Simplify:
5x - 4y - 8 = 5 - 16
5x - 4y + 3 = 0