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

$$\begin{gathered} line \frac{\mathrm{x}-\alpha }{l}=\frac{\mathrm{y}-\beta }{m}=\frac{\mathrm{z}-\gamma }{n}lies in the plane ax+by+cz+d=0 \\ then al+bm+cn=0 and a\alpha +b\beta +c\gamma +d=0 \end{gathered}$$
$$\begin{gathered} \therefore 3\left(1\right)-5\left(3\right)+2(-\mathrm{\alpha })=0 \\ 3-15-2\mathrm{\alpha }=0 \\ -12=2\mathrm{\alpha } \Rightarrow \mathrm{\alpha }=-6 \\ \begin{array}{rl} & \mathrm{ }(2,1,-2)\text{ lies on }\mathrm{x}+3\mathrm{y}+6\mathrm{z}+\mathrm{\beta }=0\\ \Rightarrow & 2+3-12+\mathrm{\beta }=0\Rightarrow \mathrm{\beta }=7 \\ \Rightarrow (\mathrm{\alpha },\mathrm{\beta })=(-6,7)\end{array} \end{gathered}$$