Let \( \dfrac{x-2}{3}=\dfrac{y+1}{-2}=\dfrac{z+3}{-1} \) lie on the plane \( px-qy+z=5 \), for some \( p,q\in\mathbb{R} \). The shortest distance of the plane from the origin is:
\( \sqrt{\dfrac{3}{109}} \)
\( \sqrt{\dfrac{5}{142}} \)
\( \dfrac{5}{\sqrt{71}} \)
\( \dfrac{1}{\sqrt{142}} \)