The equation of the chord of the ellipse $$\frac{x^2}{25} + \frac{y^2}{16} = 1,$$ whose mid-point is  $$(3,1)$$  is: 

The ellipse is $$\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$$. Write it in the standard form $$S = 0$$, where

$$S = \frac{x^{2}}{25} + \frac{y^{2}}{16} - 1 = 0$$.
If $$(x_1 , y_1)$$ is the mid-point of a chord of the conic $$S = 0$$, then the chord is given by $$T = S_1$$, where

$$T$$ is obtained from $$S$$ by replacing $$x^{2}$$ with $$x x_1$$ and $$y^{2}$$ with $$y y_1$$, and $$S_1$$ is obtained by replacing $$x, y$$ in $$S$$ with $$x_1 , y_1$$.

For our ellipse and mid-point $$(3,1)$$:

$$T = \frac{x\,x_1}{25} + \frac{y\,y_1}{16} - 1 = \frac{3x}{25} + \frac{1\;y}{16} - 1$$.

$$S_1 = \frac{x_1^{2}}{25} + \frac{y_1^{2}}{16} - 1 = \frac{3^{2}}{25} + \frac{1^{2}}{16} - 1 = \frac{9}{25} + \frac{1}{16} - 1$$.

Take the common denominator $$400$$: $$\frac{9}{25} = \frac{144}{400}, \quad \frac{1}{16} = \frac{25}{400}.$$ Hence

$$S_1 = \frac{144 + 25}{400} - 1 = \frac{169}{400} - 1 = -\frac{231}{400}.$$

Equating $$T$$ and $$S_1$$:

$$\frac{3x}{25} + \frac{y}{16} - 1 = -\frac{231}{400}.$$

Multiply every term by $$400$$ to remove fractions:

$$400 \left(\frac{3x}{25}\right) + 400 \left(\frac{y}{16}\right) - 400 = -231.$$ $$48x + 25y - 400 = -231.$$

$$48x + 25y - 400 + 231 = 0$$ $$48x + 25y - 169 = 0.$$

Therefore, the chord with mid-point $$(3,1)$$ is

$$48x + 25y = 169.$$ Hence, Option A is correct.