If the equation of the hyperbola with foci $$(4, 2)$$ and $$(8, 2)$$ is $$3x^2 - y^2 - \alpha x + \beta y + \gamma = 0$$, then $$\alpha + \beta + \gamma$$ is equal to ________.

The two foci are $$(4,2)$$ and $$(8,2)$$, so the transverse axis is horizontal and its midpoint is the centre of the hyperbola.

Centre $$\,(h,k)=\left(\frac{4+8}{2},\,2\right)=(6,2)$$.

Let the hyperbola in standard form be$$(x-h)^2/a^2-\,(y-k)^2/b^2=1.$$ Here $$c$$ is the focal distance from the centre, given by $$c=\lvert8-6\rvert=2\quad\Rightarrow\quad c^2=4.$$ For a rectangular hyperbola,$$c^2=a^2+b^2\quad-(1).$$

After shifting the origin to $$(6,2)$$ and clearing denominators we obtain $$b^2\,(x-6)^2-a^2\,(y-2)^2=a^2b^2\quad-(2).$$ Expanding $$(2)$$ gives $$b^2(x^2-12x+36)-a^2(y^2-4y+4)=a^2b^2.$$ Collect the terms on the left: $$b^2x^2-12b^2x+36b^2-a^2y^2+4a^2y-4a^2-a^2b^2=0.$$ The required general form is $$3x^2-y^2-\alpha x+\beta y+\gamma=0,$$ so we match coefficients.

Coefficient of $$x^2:$$ $$b^2=3.$$ Coefficient of $$y^2:$$ $$-a^2=-1\;\Rightarrow\;a^2=1.$$ With $$a^2=1,\;b^2=3,$$ equation $$-(1)$$ gives $$c^2=a^2+b^2=1+3=4,$$ agreeing with $$c=2,$$ so the values are consistent.

Substituting $$a^2=1,\;b^2=3$$ in $$b^2(x-6)^2-a^2(y-2)^2=a^2b^2$$: $$3(x-6)^2-(y-2)^2=3.$$ Expand: $$3(x^2-12x+36)-(y^2-4y+4)=3,$$ $$3x^2-36x+108-y^2+4y-4=3.$$ Move everything to one side: $$3x^2-y^2-36x+4y+108-4-3=0,$$ $$3x^2-y^2-36x+4y+101=0.$$ Therefore $$-\alpha=-36\;\Rightarrow\;\alpha=36,$$ $$\beta=4,$$ $$\gamma=101.$$

Finally, $$\alpha+\beta+\gamma=36+4+101=141.$$

The required value is $$141.$$