Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, intersect each other in the first quadrant. If the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is:

The two parabolas have a common vertex at the origin, with axes along the x-axis and y-axis respectively, and both have latus rectum of length 3.

For a parabola $$y^2 = 4ax$$, the latus rectum is $$4a$$. Setting $$4a = 3$$ gives $$a = \frac{3}{4}$$, so the first parabola is $$y^2 = 3x$$.

For a parabola $$x^2 = 4by$$, the latus rectum is $$4b$$. Setting $$4b = 3$$ gives $$b = \frac{3}{4}$$, so the second parabola is $$x^2 = 3y$$.

The equation of a tangent to $$y^2 = 3x$$ in slope form is $$y = mx + \frac{3}{4m}$$ (using the standard tangent $$y = mx + \frac{a}{m}$$ with $$a = \frac{3}{4}$$).

For this line to also be tangent to $$x^2 = 3y$$, substitute $$y = mx + \frac{3}{4m}$$ into $$x^2 = 3y$$: $$x^2 = 3mx + \frac{9}{4m}$$, i.e., $$x^2 - 3mx - \frac{9}{4m} = 0$$.

For tangency, the discriminant must be zero: $$\Delta = 9m^2 + 4 \cdot \frac{9}{4m} = 0$$, which gives $$9m^2 + \frac{9}{m} = 0$$.

Multiplying through by $$m$$: $$9m^3 + 9 = 0$$, so $$m^3 = -1$$, giving $$m = -1$$.

Substituting $$m = -1$$ back into the tangent equation: $$y = -x + \frac{3}{4(-1)} = -x - \frac{3}{4}$$.

Multiplying both sides by 4: $$4y = -4x - 3$$, which gives $$4x + 4y + 3 = 0$$, or equivalently $$4(x + y) + 3 = 0$$.

The correct answer is Option D: $$4(x + y) + 3 = 0$$.