If $$e_1$$ and $$e_2$$ are the eccentricities of the ellipse $$\frac{x^2}{18} + \frac{y^2}{4} = 1$$ and the hyperbola $$\frac{x^2}{9} - \frac{y^2}{4} = 1$$ respectively and $$(e_1, e_2)$$ is a point on the ellipse $$15x^2 + 3y^2 = k$$, then the value of $$k$$ is equal to:

First, we note that for every conic of the form $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ (ellipse with the major axis along the $$x$$-axis) the eccentricity formula is $$e=\sqrt{1-\frac{b^{2}}{a^{2}}}$$.

The given ellipse is $$\frac{x^{2}}{18}+\frac{y^{2}}{4}=1$$, so we identify $$a^{2}=18$$ and $$b^{2}=4$$ with $$a>b$$.

Applying the ellipse eccentricity formula, we write

$$e_{1}=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{4}{18}}.$$

Simplifying the fraction inside the square root,

$$\frac{4}{18}=\frac{2}{9},$$ so

$$e_{1}=\sqrt{1-\frac{2}{9}}=\sqrt{\frac{9}{9}-\frac{2}{9}}=\sqrt{\frac{7}{9}}.$$

Taking the square root of a fraction, we get

$$e_{1}=\frac{\sqrt7}{3}.$$

Next we consider the hyperbola $$\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$$. The standard form $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$ (transverse axis along $$x$$) has eccentricity formula $$e=\sqrt{1+\frac{b^{2}}{a^{2}}}$$. Here $$a^{2}=9$$ and $$b^{2}=4$$.

Thus

$$e_{2}=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+\frac{4}{9}}.$$

We simplify inside the root:

$$1+\frac{4}{9}=\frac{9}{9}+\frac{4}{9}=\frac{13}{9},$$ so

$$e_{2}=\sqrt{\frac{13}{9}}=\frac{\sqrt{13}}{3}.$$

Because the ordered pair $$(e_{1},e_{2})$$ lies on the ellipse $$15x^{2}+3y^{2}=k$$, we substitute $$x=e_{1}$$ and $$y=e_{2}$$ into that equation.

Hence

$$15(e_{1})^{2}+3(e_{2})^{2}=k.$$

We already have

$$e_{1}^{2}=\left(\frac{\sqrt7}{3}\right)^{2}=\frac{7}{9},\qquad e_{2}^{2}=\left(\frac{\sqrt{13}}{3}\right)^{2}=\frac{13}{9}.$$

Substituting these squared values, we obtain

$$k=15\left(\frac{7}{9}\right)+3\left(\frac{13}{9}\right).$$

Multiplying numerators and denominators,

$$15\left(\frac{7}{9}\right)=\frac{105}{9},\qquad 3\left(\frac{13}{9}\right)=\frac{39}{9}.$$

Adding the two fractions,

$$k=\frac{105}{9}+\frac{39}{9}=\frac{144}{9}.$$

Finally, dividing numerator by denominator,

$$\frac{144}{9}=16.$$

Thus $$k=16$$.

Hence, the correct answer is Option A.