If the tangent to the curve $$y = x^3$$ at the point $$P(t, t^3)$$ meets the curve again at $$Q$$, then the ordinate of the point which divides $$PQ$$ internally in the ratio 1 : 2 is:

The curve is $$y = x^3$$. At the point $$P(t, t^3)$$, the slope of the tangent is $$\frac{dy}{dx} = 3t^2$$.

The equation of the tangent at $$P$$ is $$y - t^3 = 3t^2(x - t)$$, which simplifies to $$y = 3t^2 x - 2t^3$$.

To find the other point of intersection $$Q$$, we solve $$x^3 = 3t^2 x - 2t^3$$, giving $$x^3 - 3t^2 x + 2t^3 = 0$$.

Since $$x = t$$ is a repeated root (tangent touches at $$P$$), we factor: $$(x - t)^2(x + 2t) = 0$$.

So $$Q = (-2t, (-2t)^3) = (-2t, -8t^3)$$.

The point dividing $$PQ$$ internally in the ratio $$1 : 2$$ has ordinate $$\frac{1 \times (-8t^3) + 2 \times t^3}{1 + 2} = \frac{-8t^3 + 2t^3}{3} = \frac{-6t^3}{3} = -2t^3$$.

Hence, the correct answer is Option B.