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

Given line, \( \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6} \rightarrow(1) \)
and point \( \mathrm{P}(-2,4,-5) \)
Let \( \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}=\lambda \)
So, point on line is \( Q(3 \lambda-3,5 \lambda+4,6 \lambda-8) \)

Since, \( P Q \) line is perpendicular to line of Eq. (1). SO,
\[
\begin{array}{l}
l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}=0 \\
\Rightarrow(3 \lambda-1) 3+5 \lambda(5+0)+(6 \lambda-3) 6=0 \\
\Rightarrow 9 \lambda-3+25 \lambda+36 \lambda-18=0 \\
\Rightarrow 70 \lambda=21 \Rightarrow \lambda=\frac{3}{10}
\end{array}
\]
Therefore, \( Q\left(-\frac{21}{10}, \frac{55}{10},-\frac{62}{10}\right) \)
So,
\[
\begin{array}{l}
P Q=\sqrt{\left(-\frac{21}{10}-(-2)\right)^{2}+\left(\frac{55}{10}-4\right)^{2}+\left(-\frac{62}{10}-(-5)\right)^{2}} \\
=\sqrt{\frac{1}{100}+\frac{225}{100}+\frac{144}{100}}=\sqrt{\frac{370}{100}}=\sqrt{\frac{37}{10}}
\end{array}
\]