The length of the chord of the ellipse $$\frac{x^2}{25} + \frac{y^2}{16} = 1$$, whose mid point is $$(1, \frac{2}{5})$$, is equal to:

For the ellipse $$\frac{x^2}{25} + \frac{y^2}{16} = 1$$, we need the chord with midpoint $$(1, \frac{2}{5})$$.

The equation of the chord with midpoint $$(h, k)$$ is given by $$T = S_1$$:

$$\frac{xh}{25} + \frac{yk}{16} = \frac{h^2}{25} + \frac{k^2}{16}$$

With $$(h, k) = (1, \frac{2}{5})$$:

$$\frac{x}{25} + \frac{2y/5}{16} = \frac{1}{25} + \frac{4/25}{16} = \frac{1}{25} + \frac{1}{100}$$

$$\frac{x}{25} + \frac{y}{40} = \frac{4 + 1}{100} = \frac{1}{20}$$

Multiply by 200:

$$8x + 5y = 10$$

So $$y = \frac{10 - 8x}{5} = 2 - \frac{8x}{5}$$.

Substituting into the ellipse equation:

$$\frac{x^2}{25} + \frac{(2 - 8x/5)^2}{16} = 1$$

$$\frac{x^2}{25} + \frac{4 - 32x/5 + 64x^2/25}{16} = 1$$

$$\frac{x^2}{25} + \frac{1}{4} - \frac{2x}{5} + \frac{4x^2}{25} = 1$$

$$\frac{5x^2}{25} - \frac{2x}{5} + \frac{1}{4} = 1$$

$$\frac{x^2}{5} - \frac{2x}{5} = \frac{3}{4}$$

$$\frac{x^2 - 2x}{5} = \frac{3}{4}$$

$$4(x^2 - 2x) = 15$$

$$4x^2 - 8x - 15 = 0$$

$$x = \frac{8 \pm \sqrt{64 + 240}}{8} = \frac{8 \pm \sqrt{304}}{8}$$

The two x-values: $$x_1 + x_2 = 2$$, $$x_1 x_2 = -\frac{15}{4}$$.

$$(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2 = 4 + 15 = 19$$

The corresponding y-values: $$y = 2 - \frac{8x}{5}$$, so $$y_1 - y_2 = -\frac{8}{5}(x_1 - x_2)$$.

Length of chord:

$$L^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 = (x_1 - x_2)^2\left(1 + \frac{64}{25}\right) = 19 \times \frac{89}{25} = \frac{1691}{25}$$

$$L = \frac{\sqrt{1691}}{5}$$

The answer is $$\frac{\sqrt{1691}}{5}$$, which corresponds to Option (1).