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

$\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}\to \left(1\right)$

$A\left(x_1,y_1,z_1\right)=\left(2,1,-2\right)$

$\left(l,m,n\right)=\left(3,-5,2\right)$  & $x+3y-\alpha z+\beta =0\to \left(2\right)$

$a=1,b=3,c=-\alpha$

Line (1) lines on the plane (2)

$\left(1\right)al+bm+cn=0$

$\left(II\right)a{x}_1+b{y}_1+c{z}_1+d=0$

$\Rightarrow 3-15-2\alpha =0$     $\Rightarrow \beta =7$

$\Rightarrow -12-2\alpha =0$

$\alpha =-6$