If the angle between the plane x-3y+2z= 1 and their $\frac{x-1}{2}=\frac{y-1}{1}=\frac{z-1}{-3}$ is $\theta$ then find the value of $\mathrm{cosec} \mathrm{\theta }$

Vector normal to the plane is $\overrightarrow{\mathrm{n}}=\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}+2\widehat{\mathrm{k}}$  

and vector along the line is  $\overrightarrow{\mathrm{v}}=2\widehat{\mathrm{i}}+\widehat{\mathrm{j}}-3\widehat{\mathrm{k}}$

Now $\sin \theta =\frac{\overrightarrow{x}\cdot \overrightarrow{v}}{\left|\overrightarrow{x}\right|\left|\overrightarrow{v}\right|}=\left|\frac{2-3-6}{\sqrt{14}\sqrt{14}}\right|=\left|\frac{7}{14}\right|$

Hence, $\mathrm{cosec} \mathrm{\theta }=2$