1. Identify the center and transverse axis

The foci are given as $(4,2)$ and $(8,2)$.

• Since the foci have the same $y$-coordinate, the hyperbola opens horizontally.
• The center is the midpoint of the foci: $\left(\frac{4+8}{2},\frac{2+2}{2}\right)=(6,2).$

So the standard form is $\frac{(x-6)^2}{a^2}-\frac{(y-2)^2}{b^2}=1.$

2. Use the focal distance

For a horizontal hyperbola, $c^2=a^2+b^2,$ where the foci are $(h\pm c,k)$.

Here, $c=2$ because the foci are $2$ units away from the center $(6,2)$.

Thus, $a^2+b^2=4.$

3. Use the given quadratic form

The equation is given as $3x^2-y^2-\alpha x+\beta y+\gamma=0.$

Notice the coefficients of $x^2$ and $y^2$ are $3$ and $-1$. So the hyperbola should be of the form $3(x-6)^2-(y-2)^2=\text{constant}.$

Let us expand the standard form more carefully.

From $\frac{(x-6)^2}{a^2}-\frac{(y-2)^2}{b^2}=1,$ we get $b^2(x-6)^2-a^2(y-2)^2=a^2b^2.$

Comparing with the required coefficients $3x^2-y^2$, we need $b^2=3,\qquad a^2=1.$

Then indeed, $a^2+b^2=1+3=4,$ which matches $c^2=4$. So this is consistent.

Hence the hyperbola is $\frac{(x-6)^2}{1}-\frac{(y-2)^2}{3}=1.$

Multiplying by $3$: $3(x-6)^2-(y-2)^2=3.$

4. Expand the equation

Expand: $3(x^2-12x+36)-(y^2-4y+4)=3.$

So, $3x^2-36x+108-y^2+4y-4=3.$

Bring all terms to one side: $3x^2-y^2-36x+4y+101=0.$

Comparing with $3x^2-y^2-\alpha x+\beta y+\gamma=0,$ we get $\alpha=36,\quad \beta=4,\quad \gamma=101.$

5. Find the required sum

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