The distance of the point \(\mathrm{P}(-2,4,-5)\) from the line \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\) is

Since the point is \((-2,4,-5)\),
\(
\therefore \mathrm{a}=-2, \mathrm{~b}=4, \mathrm{c}=-5
\)
Given equation of line is
\(
\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}
\)
\(
\therefore \quad x_1=-3, y_1=4, z_1=-8
\)
d.r.s of the line are \(3,5,6\)
\(\therefore\) d.c.s are \(\frac{3}{\sqrt{70}}, \frac{5}{\sqrt{70}}, \frac{6}{\sqrt{70}}\)
Perpendicular distance of point from the line is
\(
\begin{aligned}
& \sqrt{\begin{array}{r}
{\left[\left(\mathrm{a}-x_1\right)^2+\left(\mathrm{b}-y_1\right)^2+\left(\mathrm{c}-\mathrm{z}_1\right)^2\right]} \\
-\left[\left(\mathrm{a}-x_1\right) l+\left(\mathrm{b}-y_1\right) \mathrm{m}+\left(\mathrm{c}-\mathrm{z}_1\right) \mathrm{n}\right]^2
\end{array}} \\
= & \sqrt{1^2+0+3^2-\left[\frac{1(3)}{\sqrt{70}}+\frac{0(5)}{\sqrt{70}}+\frac{3(6)}{\sqrt{70}}\right]^2} \\
= & \sqrt{1+9-\left(\frac{3}{\sqrt{70}}+\frac{18}{\sqrt{70}}\right)^2} \\
= & \sqrt{\frac{37}{10}} \text { units }
\end{aligned}
\)