The distance of point A(-2, 3, 1) from the line PQ through P(-3, 5, 2), which make equal angles with the axes is

$$\begin{gathered} \mathrm{Here} \mathrm{\alpha }=\mathrm{\beta }=\mathrm{\gamma } \mathrm{Since} {\cos }^2\mathrm{\alpha }+{\cos }^2\mathrm{\beta }+{\cos }^2\mathrm{\gamma }=1, \cos \mathrm{\alpha }=1/\sqrt{3} \\ \mathrm{DC}\text{'}\mathrm{s} \mathrm{of} \mathrm{PQ} \mathrm{are} \left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right) \end{gathered}$$

$$\begin{gathered} =\left|\left(-2+3\right)\frac{1}{\sqrt{3}}+\left(3-5\right)\frac{1}{\sqrt{3}}+\left(1-2\right)\frac{1}{\sqrt{3}}\right|=\frac{2}{\sqrt{3}} \\ \mathrm{AP}=\sqrt{{\left(-2+3\right)}^2+{\left(3-5\right)}^2{\left(1-2\right)}^2}=\sqrt{6} \\ \mathrm{AM}=\sqrt{{\left(\mathrm{AP}\right)}^2-{\left(\mathrm{PM}\right)}^2}=\sqrt{6-\frac{4}{3}}=\sqrt{\frac{14}{3}} \end{gathered}$$