Step-by-step Solution:

We are given the circle \[ x^{2}+y^{2}-6x-10y+30=0. \] Step 1: Centre and radius. Complete the squares: \[ (x-3)^{2}+(y-5)^{2}=4. \] Hence the centre is \(C(3,5)\) and the radius is \(2\). Step 2: Reflection of the centre across the line. The mirror line is \[ 3x+y=2 \quad \Leftrightarrow \quad 3x+y-2=0. \] Here \(a=3,\;b=1,\;c=-2\). For a point \((x_0,y_0)\), its reflection across \(ax+by+c=0\) is \[ x'=x_0-\frac{2a(ax_0+by_0+c)}{a^2+b^2}, \quad y'=y_0-\frac{2b(ax_0+by_0+c)}{a^2+b^2}. \] For \(C(3,5)\), \[ ax_0+by_0+c = 3\cdot 3 + 1\cdot 5 - 2 = 12, \qquad a^2+b^2=9+1=10. \] Thus \[ x' = 3 - \frac{2\cdot 3 \cdot 12}{10} = 3 - \frac{72}{10} = 3 - \frac{36}{5} = -\frac{21}{5}, \] \[ y' = 5 - \frac{2 \cdot 1 \cdot 12}{10} = 5 - \frac{24}{10} = 5 - \frac{12}{5} = \frac{13}{5}. \] So the reflected centre is \[ C'\left(-\tfrac{21}{5}, \tfrac{13}{5}\right). \] Step 3: Equation of the image circle. The radius remains \(2\). Hence, \[ \left(x+\tfrac{21}{5}\right)^{2}+\left(y-\tfrac{13}{5}\right)^{2}=4. \] Expanding: \[ x^{2}+y^{2}+\tfrac{42}{5}x-\tfrac{26}{5}y+\frac{441+169}{25}-4=0. \] Simplify constant term: \[ \frac{441+169}{25}-4=\frac{610}{25}-\frac{100}{25}=\frac{510}{25}=\frac{102}{5}. \] Thus the equation is \[ x^{2}+y^{2}+\tfrac{42}{5}x-\tfrac{26}{5}y+\tfrac{102}{5}=0. \] Step 4: Compute \(\alpha+\beta+\gamma\). Here \[ \alpha=\tfrac{42}{5},\;\;\beta=-\tfrac{26}{5},\;\;\gamma=\tfrac{102}{5}. \] So \[ \alpha+\beta+\gamma = \frac{42-26+102}{5} = \frac{118}{5} = 23.6. \] Therefore, \[ \lfloor \alpha+\beta+\gamma \rfloor = \boxed{23}. \]