$$\text{If the equation of the parabola with vertex }V\left(\frac{3}{2},3\right)\text{ and the directrix } x+2y=0\text{ is }\alpha x^2+\beta y^2-\gamma xy-30x-60y+225=0,\text{then } \alpha+\beta+\gamma \text{ is equal to:}$$

We need to find $$\alpha + \beta + \gamma$$ for the parabola with given vertex and directrix.

The vertex is $$V\left(\frac{3}{2}, 3\right)$$ and the directrix is $$x + 2y = 0$$.

Since the axis of the parabola is perpendicular to the directrix and passes through the vertex, and because the directrix $$x + 2y = 0$$ has slope $$-1/2$$, it follows that the axis has slope 2, i.e., direction vector $$(1,2)$$.

Next, the distance from the vertex to the directrix is given by $$d = \frac{|3/2 + 2(3)|}{\sqrt{1^2 + 2^2}} = \frac{|3/2 + 6|}{\sqrt{5}} = \frac{15/2}{\sqrt{5}} = \frac{15}{2\sqrt{5}}$$.

Since the focus is on the axis, on the opposite side of the vertex from the directrix, at the same distance $$d$$, the unit vector along the axis (away from the directrix) is $$\frac{(1,2)}{\sqrt{5}}$$, so

$$F = V + d \cdot \frac{(1,2)}{\sqrt{5}} = \left(\frac{3}{2}, 3\right) + \frac{15}{2\sqrt{5}} \cdot \frac{(1,2)}{\sqrt{5}} = \left(\frac{3}{2}, 3\right) + \frac{15}{10}(1,2) = \left(\frac{3}{2}+\frac{3}{2}, 3+3\right) = (3, 6)$$

Using the definition of a parabola as the locus of points equidistant from the focus and directrix leads to

$$\sqrt{(x-3)^2 + (y-6)^2} = \frac{|x+2y|}{\sqrt{5}}$$

Squaring both sides gives

$$ (x-3)^2 + (y-6)^2 = \frac{(x+2y)^2}{5} $$

and hence

$$ 5[(x-3)^2 + (y-6)^2] = (x+2y)^2. $$

Expanding this equation results in

$$5(x^2 - 6x + 9 + y^2 - 12y + 36) = x^2 + 4xy + 4y^2$$

which simplifies to

$$5x^2 - 30x + 45 + 5y^2 - 60y + 180 = x^2 + 4xy + 4y^2$$

and therefore

$$4x^2 + y^2 - 4xy - 30x - 60y + 225 = 0$$.

Comparing this with the given form $$\alpha x^2 + \beta y^2 - \gamma xy - 30x - 60y + 225 = 0$$ shows that $$\alpha = 4$$, $$\beta = 1$$, and $$\gamma = 4$$, so $$\alpha + \beta + \gamma = 4 + 1 + 4 = 9$$.

The correct answer is Option 2: 9.