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

(N/A) Let the given point be $P(-2, 4, -5)$.
Any point $Q$ on the line $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} = \lambda$ is given by $Q(3\lambda - 3, 5\lambda + 4, 6\lambda - 8)$.
The vector $\overrightarrow{PQ}$ is given by $\overrightarrow{PQ} = (3\lambda - 3 - (-2))\hat{i} + (5\lambda + 4 - 4)\hat{j} + (6\lambda - 8 - (-5))\hat{k} = (3\lambda - 1)\hat{i} + 5\lambda\hat{j} + (6\lambda - 3)\hat{k}$.
Since $\overrightarrow{PQ}$ is perpendicular to the line with direction vector $\vec{v} = 3\hat{i} + 5\hat{j} + 6\hat{k}$,their dot product must be zero:
$\overrightarrow{PQ} \cdot \vec{v} = 0$
$3(3\lambda - 1) + 5(5\lambda) + 6(6\lambda - 3) = 0$
$9\lambda - 3 + 25\lambda + 36\lambda - 18 = 0$
$70\lambda - 21 = 0 \implies \lambda = \frac{21}{70} = \frac{3}{10}$.
Substituting $\lambda = \frac{3}{10}$ into $\overrightarrow{PQ}$:
$\overrightarrow{PQ} = (3(\frac{3}{10}) - 1)\hat{i} + 5(\frac{3}{10})\hat{j} + (6(\frac{3}{10}) - 3)\hat{k} = (\frac{9-10}{10})\hat{i} + \frac{15}{10}\hat{j} + (\frac{18-30}{10})\hat{k} = -\frac{1}{10}\hat{i} + \frac{15}{10}\hat{j} - \frac{12}{10}\hat{k}$.
The distance is the magnitude of $\overrightarrow{PQ}$:
$|\overrightarrow{PQ}| = \sqrt{(-\frac{1}{10})^2 + (\frac{15}{10})^2 + (-\frac{12}{10})^2} = \sqrt{\frac{1 + 225 + 144}{100}} = \sqrt{\frac{370}{100}} = \sqrt{\frac{37}{10}} \text{ units.}$