Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$lies in the plane $x+3y-\alpha z+\beta =0$. Then $\left(\alpha ,\beta \right)$ equals 

The given line is $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$

The given plane is $x+3y-\alpha z+\beta =0$

The condition that the line $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$ lies on the plane $ax+by+cz+d=0$ is $al+bm+cn=0$ and $a{x}_1+b{y}_1+c{z}_1+d=0$

hence, 

   $\begin{array}{rcl}3-15-2\alpha & =& 0\\ 2\alpha & =& 12\\ \alpha & =& 6\end{array}$

The point $\left(2,1,-2\right)$ lies on the plane $x+3y-\alpha z+\beta =0$

it implies that $\begin{array}{rcl}2+3+2\alpha +\beta & =& 0\\ 5+12+\beta & =& 0\\ \beta & =& -17\end{array}$

Therefore, $\left(\alpha ,\beta \right)=\boxed{\left(6,-17\right)}$