$\text{ The distance of the point }-\overline{i}+2\overline{j}+6\overline{k}\text{ from the straight line through the point }(2,3,-4)\text{ and }$$\text{ parallel to }6\overline{i}+3\overline{j}-4\overline{k}\text{ is }$

[SolutionStep1]:

From the figure

$\text{ Dr's of }\overrightarrow{AP}=(-3,-1,10) \left(x_2-x_1,y_2-y_1,z_2-z_1\right)$

[SolutionStep2]:

$\left|\overrightarrow{AP}\right|=\sqrt{110}$

$\overline{PN}\text{ vector is perpendicular to the line }$

[SolutionStep3]:

$\overrightarrow{AN}=\text{ The projection of }\overrightarrow{AP}\text{ one the line }(on(6,3,-4\left)\right)$

$=\left|\frac{\overrightarrow{AP}.\left(6,3,-4\right)}{\left(6,3,-4\right)}\right|$

$=\left|\frac{\left(-3,-1,10\right).\left(6,3,-4\right)}{\left(6,3,-4\right)}\right|$

$=\frac{\left|-18-3-40\right|}{\sqrt{61}}=\sqrt{61}$

$\text{ From figure }P{N}^2=A{P}^2-A{N}^2$

$=110-61=49$

$\Rightarrow PN=7$