Point P(0, 1,−2) to the line given by $r\left(t\right)=\left(1+5t,-6+t,-4t\right)$

A point $P(0,1,-2)$ to the line by $r\left(t\right)=(1+5t,-6+t,-4t)$

The goal is to calculate the distance between the point and the line.

The formula for the distance is $\frac{||d\times {\overrightarrow{P}}_{0}P||}{\Vert d\Vert }$

Let the point $P$is $P(0,1,-2)$

The point $P_0$ on the line equation $r\left(t\right)=(1+5t,-6+t,-4t)$ is $(1,-6,0)$

The direction vector $\overrightarrow{P_{0}P}=(0-1,1-(-6),-2-0)$

Then $\overrightarrow{P_{0}P}=(-1,7,-2)$

Take the direction vector of the equation $r\left(t\right)=(1+5t,-6+t,-4t)$is $d=(5,1,-4)$

Substitute the values $d=(5,1,-4)$and $\overrightarrow{P_{0}P}=(-1,7,-2)$in the formula $\underset{¯}{\Vert d\times \overrightarrow{P_{0}P}}$

Then the distance $=\frac{\Vert (5,1,-4)\times (-1,7,-2)\Vert }{\Vert (5,1,-4)\Vert }$

The cross product of $(5,1,-4)\times (-1,7,-2)$ is calculated as follows,

$$\begin{gathered} =i(-2+28)-j(-10-4)+k(35+1) \\ =26i+14j+36k \end{gathered}$$

Thus,

On further simplification,

Distance $=\sqrt{\frac{1084}{21}}$ units

Thus, the distance between the point $P(0,1,-2)$ and the line $r\left(t\right)=(1+5t,-6+t,-4t)$ is $\sqrt{\frac{1084}{21}}$ units.

Therefore, the answer is $\sqrt{\frac{1084}{21}}$ units.